Introductory probability - Hawker Maths 202011E Karnaugh maps and probability tables 11F Conditional...

ChapTEr 11 • Introductory probability 475

ChapTEr ConTEnTS 11a Introduction to experimental probability 11B Calculating probabilities 11C Tree diagrams and lattice diagrams 11d The Addition Law of Probabilities 11E Karnaugh maps and probability tables 11F Conditional probability 11G Transition matrices and Markov chains 11h Independent events 11i Simulation

diGiTal doCdoc-980110 Quick Questions

ChapTEr 11

Introductory probability

11a introduction to experimental probabilityTossing a fair coin or rolling a standard 6-sided die will result in a range of outcomes. The coin can land Heads or Tails, and the number appearing uppermost on the die will be one of the numbers 1, 2, 3, 4, 5 or 6. Probability involves assigning a numerical value to the likelihood of such events occurring.

In this respect, certain events will clearly be more probable than others; for example, getting only 1 of the required 6 numbers in Tattslotto is more likely than obtaining all 6 winning numbers.

A numerical value for the probability of an event can be established in a number of ways. It can be based on results arising from experiments; alternatively, a reasoned estimate of the likelihood of the event can be provided on the strength of personal experience and knowledge (the subjective probability). A third way is to consider the ‘symmetry’ of the situation where the activity has equiprobable or equally likely outcomes. For example, if we toss a coin 50 times and note how many times it lands ‘Heads’ (a Head facing up), we may conclude (based on the experiment) that the probability of a coin landing Heads up is half. We may also reason that a tossed coin has two equally likely outcomes (a Head and a Tail), of which Heads is one possibility, so there is 1 chance in 2, or 1

2,

of a Head. However, deciding what the chances are of a runner winning her race will be subjective and dependent on considerations such as the runner’s past performances, her current state of fi tness and the abilities of the other competitors.

random outcome experimentsWhat is the probability of a fair coin landing Tails? For a single trial of this experiment (one toss of the coin), we know the coin will land either Heads or Tails, but we cannot be sure the toss will produce a favourable outcome (that is, a Tail). The result is a random outcome. (Closing one’s eyes and taking out a marble from a box containing different coloured marbles, or shuffl ing a pack of playing cards and choosing the topmost card, are also activities that produce random outcomes.)

For our example of the coin, if many trials are conducted we will observe that the ratio number of Tails

total number of trials, which is the experimental probability for the favourable outcomes, converges

(‘gets closer’) to a particular value. This particular value is known as the long-run proportion.

diGiTal doCSdoc-9802long run proportiondoc-9803one diedoc-9804Two dice

ConTEnTS

11a introduction to experimental probability 475 Exercise 11a introduction to experimental probability 477

11B Calculating probabilities 478 Exercise 11B Calculating probabilities 481

11C Tree diagrams and lattice diagrams 483 Exercise 11C Tree diagrams and lattice diagrams 485

11d The addition law of probabilities 487 Exercise 11d The addition law of probabilities 490

11E karnaugh maps and probability tables 492 Exercise 11E karnaugh maps and probability tables 495

476 Maths Quest 11 Mathematical Methods CAS

Similarly, we observe that the ratio number of Heads

total number of trials, the experimental probability for a Head,

converges to a particular value.For a coin tossed many times the long-run proportion of a Head is 0.5 and the long-run proportion of a

Tail is 0.5.

Experimental probability and expected number of outcomesIn general, the experimental probability is given by:

experimental probability = number of favourable outcomes observed

total number of trials

The number of times an outcome of an activity is expected to occur is given by:

expected number of favourable outcomes = experimental probability (long-run proportion) × number of trials

WorkEd ExamplE 1

A 6-sided die (not necessarily a fair one) was rolled 12 times and the number showing uppermost was noted each time. The numbers uppermost on the die were:

2, 4, 1, 1, 5, 6, 4, 3, 4, 5, 6, 1.Estimate the probability of rolling a 5 with this die.

Think WriTE

1 There are 2 favourable outcomes.

2 There are 12 outcomes altogether.

3 Use the formula: experimental probability

= number of favourable outcomes observed


Experimental probability = 212

= 16

WorkEd ExamplE 2

A fair 6-sided die is rolled 48 times. How many times is an even number expected to show uppermost?

Think WriTE

1 There are 6 equally likely outcomes for the roll of the die and 3 favourable outcomes corresponding to an even number.

Experimental probability of an even number

= 36

= 12

2 There are 48 trials. Use the formula expected number of favourable outcomes = experimental probability × number of trials.

Expected number of even numbers

= 12 × 48

= 24

WorkEd ExamplE 3

Inside a bag are 18 marbles, some white and the rest green. One marble is taken out without looking, its colour is noted and the marble put back inside the bag. When this is done 30 times it is found that a green marble was taken out 5 times. Estimate how many marbles of each colour are in the bag.


Think WriTE

1 A green marble was taken out 5 times and a white marble 25 times. Work out the experimental probabilities.

Experimental probability of green marble = 530

= 16

Experimental probability of white marble = 2530

= 56

2 Calculate the expected number of each colour marble. Use the formula expected number of favourable outcomes = experimental probability × number of trials.

Expected number of green marbles = 16 × 18

= 3

Expected number of white marbles = 56 × 18

= 15Estimated number of each type of marble: 3 green, 15 white

Exercise 11a introduction to experimental probability1 WE 1 A coin was tossed 10 times and the outcomes noted as H, T, H, H, T, T, T, T, H, T, where H is a

Head and T is a Tail. Find the experimental probability of a Tail.

2 Twenty letters were chosen at random from the alphabet and recorded as either consonant, c, or vowel, v. The results were c, c, v, c, v, c, c, c, c, v, c, v, v, c, c, c, c, v, c, c. Calculate the experimental probability of choosing a consonant.

3 A biased coin is tossed 50 times. The results were 33 Tails and 17 Heads.a What is the experimental probability of tossing a Tail with this coin?b What is the experimental probability of tossing a Head with this coin?

4 WE2 A die is tossed 96 times. How many times is an odd number expected to appear uppermost on the die?

5 A coin is tossed 500 times. What is the expected number of Heads?

6 A die is rolled 300 times. How many odd numbers or the number 2 are expected to turn up?

7 mC A die is tossed 102 times. The number of times a number between 1 and 3 inclusive is expected to appear uppermost on the die is:a 51 B 34 C 20 d 64 E 68

8 mC A box contains 2 blue beads, 3 green beads and 1 yellow bead. One bead is taken out, its colour is noted and it is put back in the box. This is repeated 246 times. The number of times a bead that is not yellow is expected to be taken out of the box is:a 41 B 82 C 205 d 123 E 164

9 WE3 Inside a box are 42 plastic shapes. Some of the shapes are squares and the remainder are circles. One shape is taken out at random, its shape is noted and it is put back in the box. After this is repeated 84 times it is found that a square was taken out 36 times. Estimate how many squares and how many circles are in the bag.

10 A closed box with a hole in one corner contains coloured marbles: 4 are red, 2 are blue, 3 are white and 1 is green. The box is shaken and 1 marble falls out. Its colour is recorded and it is placed back in the box. This is done 200 times.a How many times is a red or blue marble expected to fall out of the box after 200 trials?b How many times is a marble that has a colour other than white expected to fall out of the box after

200 trials?

11 mC A moneybox contains 128 coins. There are 5-cent and 10-cent coins. The box is shaken, a coin falls out, the value of the coin is noted and it is placed back inside the box. After this is repeated 96 times it is noted that a 5-cent coin fell out 60 times. The estimated number of 10-cent coins in the moneybox is:a 24 B 64 C 60 d 36 E 48


12 During a period of one week 190 people telephoned Hot-Shot Electrics with enquiries. During the same period Zap Inc received 305 enquiries. Based on this information, how many enquiries did Hot-Shot and Zap Inc each expect during a week where the total number of phone calls made to the two businesses was 650 (to the nearest call)?

13 The probability that the Last Legs football team can win a match is 2

7. If the team is to play 35 matches during the season, how

many wins should it expect?

14 In a 9-game chess tournament, Adam won 6 games, lost 2 games and drew 1 game.a Based on this information, if Adam is to play 108 games next

year, how many games should he expect to: i win? ii lose? iii draw?b Based on the fact that Adam won 81 of the 108 games, how

many games does he expect to lose or to draw in a tournament comprising 16 games?

15 mC Inside a bag are red, blue and black marbles. Sally takes out one marble, notes its colour and puts it back in the bag. When she has taken out a marble 360 times, she finds that a red marble was taken out 140 times and a black marble 200 times. If Sally takes out a marble 270 times, the number of blue marbles expected is:a 15 B 90 C 105 d 150 E 125

16 A post-office has two letterboxes, Domestic and Overseas. Letters to be delivered within Australia are placed in the Domestic box, and letters intended for overseas destinations are deposited in the Overseas box. During the month of January there were 980 Domestic and 310 Overseas letters handled.a Estimate the probability of a particular letter having an Australian destination.b Estimate the probability of a particular letter having an overseas destination.c During February there were 1580 letters posted in total. How many of these would you expect to

have been delivered: i within Australia? ii overseas?

11B Calculating probabilitiesMany of the methods arising from a study of probability can be investigated by using set theory. A review of the basic work on sets is provided in your eBookPLUS.

This section describes how to calculate exact theoretical probabilities rather than use experimental results to estimate probabilities. We know that the theoretical probability of a fair coin coming up Heads is exactly 1

2.

We must remember that this does not mean that exactly half the tosses of the coin will be Heads, but rather that the long-run proportion of Heads will approach 1

2 as the number of tosses becomes very large.

If n is the number of trials, then as n → ∞ (→ means ‘approaches’, or ‘gets closer to’): the proportion of successes → the theoretical probability of a ‘success’.

Before describing how to calculate theoretical probability (and avoid the need to perform a large number of trials), we need to discuss the ideas of event space and favourable outcome in more detail.

Event spaceThe event space (or sample space) consists of all possible outcomes of an experiment. The event space is the universal set and is denoted by ξ.

WorkEd ExamplE 4

A 6-sided die is rolled. List the elements of the event space and state the cardinal number.

Think WriTE

1 List the elements of the event space. ξ = {1, 2, 3, 4, 5, 6}

2 Count the number of elements in the event space. n(ξ) = 6

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WorkSHEET 11.1

diGiTal doCdoc-9810Extension

Sets and Venn diagrams


WorkEd ExamplE 5

A coin and a die are tossed.a List the elements of the event space.b List the elements of the event E = ‘Head and a number greater than 4’.

Think WriTE

a What are the different outcomes using a coin and a die together?

a Let H be Head, T be Tail and H4 mean ‘Head on the coin and a 4 on the die’. Then ξ = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}.

b List all the possible ways of obtaining event E = ‘Head and a number greater than 4’.

b Event E consists of 2 sample points: E = {H5, H6}.

probabilitiesThe game of ‘Zilch’ involves tossing a fair 6-sided die and scoring points for rolling a 6 or a 1.

The events ‘rolling a 6’ and ‘rolling a 1’ are called ‘favourable outcomes’.The total number of outcomes is 6 (a result of 1, 2, 3, 4, 5 or 6 could be rolled). Each outcome is

equally likely for a fair die.Intuition may lead us to assert that the probability of scoring by rolling a die in a game of Zilch is 2

6 = 1

3.

More formally, for equally likely outcomes:

probability of a favourable outcome = number of favourable outcomes

total number of possible outcome

or

Pr(favourable outcome) = number of favourable outcomes

total number of possible outcomeSo in our Zilch example,

Pr(scoring) = number of favourable outcomes

total number of possible outcome Pr(scoring) = 2

6

= 13 as before.

There are other, equivalent expressions for calculating probability, including

Pr(E) = number of favourable outcomes in E

total number of possible outcomes

Or, using set notation,

Pr(E) = n(E )

n(ξ )where Pr(E ) is the probability of event E, n(E ) is the cardinal number of event E and n(ξ ) is the cardinal number of the event space. The above Zilch example may be illustrated as follows, where E = getting a 6 or a 1.

Notice that Pr(E ′) + Pr(E ′) = 26 + 4

6

= 1.In general, if E and E ′ are complementary events,

Pr(E) + Pr(E′) = 1and

Pr(E′) = 1 − Pr(E)

ξE

62

3

54

1


WorkEd ExamplE 6

A number is randomly chosen from the first 12 positive integers. Find the probability of:a choosing the number 8b choosing any number except 8.

Think WriTE

a 1 Pr(favourable outcome)

= number of favourable outcomes


a

2 There is one favourable outcome (choosing an 8) and 12 possible outcomes.

Pr(8) = 112

b We require the complementary probability. Pr(E ′ ) = 1 − Pr(E ).

b Pr(not 8) = 1 − 112

= 1112

range of probabilitiesIf there is no favourable outcome for event E, then n(E ) = 0, so:

Pr(E) = n(E )

n(ξ )

= 0

n(ξ )

= 0

We interpret this to mean that impossible events have a probability of zero.If every outcome in the event space for E is a favourable outcome, then

n(E ) = n(ξ) and

Pr(E) = n(E )

n(ξ )

= n(ξ )

n(ξ )

= 1.

We interpret this to mean that events certain to happen have a probability of 1. Thus the range of values for the probability of an event is given by 0 ≤ Pr(E ) ≤ 1. The probability line below illustrates the range of probabilities and the likelihood of the event occurring.

Impossible Unlikely Equallylikely

Likely Certain

1–2

1–4

3–40 1

WorkEd ExamplE 7

A fair cubic die with faces numbered 1, 3, 4, 6, 8, 10 is rolled. Determine the probability that the number appearing uppermost will be:a even b odd c less than 1 d greater than or equal to 1.

Think WriTE

a 1 There are 6 possible outcomes when rolling the die. a n(ξ ) = 6

2 Four of the outcomes correspond to an even number.

Pr(even number) = 46

= 23


b Two of the outcomes correspond to an odd number.

Also 1 − 23 = 1

3 since

Pr(odd number) + Pr(even number) = 1

b Pr(odd number) = 26

= 13

c None of the outcomes correspond to a number less than 1.

c Pr(number is less than 1) = 06

= 0

d All 6 outcomes correspond to a number greater than or equal to 1.

d Pr(number is greater than or equal to 1) = 66

= 1

In worked example 7, notice that if A is the event ‘even number’ then the complement (A′) of A is the event ‘odd number’ and

Pr(A) + Pr(A′) = 46 + 2

6

= 1.

Similarly, if B is the event ‘a number less than 1’ then B′ is the event ‘a number greater than or equal to 1’ so that

Pr(B) + Pr(B′) = 06 + 6

6

= 1

WorkEd ExamplE 8

One letter is randomly selected from the letters in the sentence LITTLE MISS MUFFETT. Calculate the probability that the letter is:a a vowelb a consonant other than a Tc a consonant.

Think WriTE

a 1 Pr(favourable outcome)

= number of favourable outcomes


a

2 A vowel is a favourable outcome. There are 17 possible outcomes (letters), of which 5 are vowels.

3 Substitute this information into the probability formula.

Pr(vowel) = 517

b 1 There are 8 consonants other than T. b

2 Use the probability formula. Pr(consonant other than T) = 817

c 1 There are 12 consonants. c

2 Write the probability. Pr(consonant) = 1217

Exercise 11B Calculating probabilities1 WE4 A spinner is divided into 4 equal sections as shown at right.

For one spin:a list the elements in the event spaceb state the cardinal number of the event space.

Blue

GreenYellow

Red


2 A numberplate is made up of 3 letters followed by 3 numbers. What is the event space for the first position on the numberplate?

3 A card is chosen from a pack of 52 playing cards. What is the cardinal number of the event space?

4 WE5 A card is chosen from a pack of 52 playing cards and its suit noted, then a coin is tossed.a List the elements in the event space.b List the elements in the event S = ‘a spade is chosen’.

5 A coin is tossed twice. List the elements in the event space.

6 A student is chosen at random from a class of 12 girls and 14 boys, then a chocolate bar is chosen from a bag containing a Time Out, a Mars Bar and a Violet Crumble.a List the elements in the event space.b List the elements in the event M = ‘a Mars Bar is chosen’.

7 WE6 One player is chosen at random from the senior netball team to be the captain. If there are 7 players in the team, what is the probability the person who plays goal attack is:a chosen? b not chosen?

8 One Year 11 student must be chosen to represent the year level at a staff meeting. If all 81 girls’ and 73 boys’ names are put into a container and one name is chosen at random, find the probability that:a a Year 11 student is chosenb any particular Year 11 student is chosenc a boy is chosen.

9 One card is chosen from a pack of 52 playing cards. What is the probability that the card is:a a queen? b a heart?c a picture card (J, Q, K)? d not a picture card?e red or black?

10 Four hundred thousand tickets are sold in a raffle. The winner of the raffle will toss the coin at the AFL grand final. If you bought 10 tickets, what is the probability that you will win?

11 WE7 A standard die is rolled. What is the probability of rolling:a an even number?b a 5?c a number from 2 to 4 inclusive?d a number less than 7?

12 A bag has 20 marbles numbered 1, 2, 3, . . . , 20. One marble is randomly drawn. Find the probability that the number on the marble is:a even b greater than 4 c a multiple of 4 d not even.

13 WE8 One letter is randomly selected from the letters of the palindrome ‘Madam, I’m Adam’. Calculate the probability that the letter is:a a vowel b a consonant other than a D.

14 What is the probability of randomly choosing a consonant other than P from the letters of the palindrome ‘A man, a plan, a canal, Panama’?

15 One letter is randomly selected from the words Mathematical Methods. What is the probability of randomly selecting:a the letter m?b a vowel?c a consonant?d a letter from the first half of the alphabet?

16 A lolly shop has 85 different types of lollies including Smarties in clear plastic containers. Forty of the lollies contain chocolate. If one container is chosen at random, what is the probability it contains:a a lolly containing chocolate?b Smarties?c a biscuit?


11C Tree diagrams and lattice diagramsTree diagramsA useful way of representing all possible outcomes for sequential activities is by means of a tree diagram. A tree diagram consists of paths formed from branches. Each sample point (possible outcome) corresponds to a unique path that is found by following the branches. For example, a tree diagram could be drawn to show all possible outcomes when a coin is tossed twice.

The fi rst set of branches shows the possible outcome of the fi rst activity, in this case tossing the coin the fi rst time.

The second set of branches is then joined onto the ends of the fi rst set to show all outcomes of both tosses of the coin.

Note that the outcomes are written at the end of each path through the tree diagram.

The cardinal number of the sample space is the total number of end branches at the end of each path. If all outcomes are equally likely, the probability can then be determined as before by using

Pr(E ) = n(E )

n(ξ ).

WorkEd ExamplE 9

A card is chosen from a pack of 52 playing cards and its suit noted; then it is returned to the pack before another card is chosen.a Draw a tree diagram showing all possible suit outcomes.b Calculate the probability of choosing: i two hearts ii a diamond then a spade iii a heart and a club.

Think WriTE/draW

a Draw a tree diagram.

S — spade

H — heart

D — diamond

C — club

a 1st card 2nd card Outcome

S

H

D

C

SSSHDCSHDCSHDCSHDC

SHSDSCHSHHHDHCDSDHDDDCCSCHCDCC

b i Use the probability formula with one favourable outcome (heart, heart) out of 16 possible outcomes.

b i Pr(HH) = 116

ii Use the probability formula with one favourable outcome (diamond, spade) out of 16 possible outcomes.

ii Pr(DS) = 116

iii Use the probability formula with two favourable outcomes (heart then club or club then heart) out of 16 possible outcomes.

iii Pr(HC or CH) = 216

= 18

1st coin

H

T

1st coin 2nd coin

H

TH

T

H

T

HH

HTTH

TT

Outcome

TUTorialeles-1448Worked example 9


WorkEd ExamplE 10

Two letters are selected from the word BIRD.a Draw a tree diagram to illustrate the event space.b What is the probability that the second letter is a vowel or that the first letter is D?

Think WriTE/draW

a 1 There are 4 letters to choose from as the first letter of the pair of letters.

a 1st letter 2nd letter Outcomes

BRD

BID

BIR

IRD

I

B

R

D

BIBRBD

IBIRID

RBRIRD

DBDIDR

2 For each letter chosen as the first letter, there are 3 letters remaining to choose from.

b There are 5 favourable outcomes {BI, RI, DI, DB and DR} and 12 outcomes altogether.

b Pr(second letter is a vowel or first letter is D)

= 512

lattice diagramsWhen showing all possible outcomes of two activities such as ‘a die is rolled twice’, a tree diagram can become very large. An alternative method of showing all possible outcomes in this situation is a lattice diagram.

A lattice diagram is a graphical representation in which the axes show the possible outcomes of each activity. The ‘coordinates’ or points inside the graph show the possible outcomes from the combination of both activities, for example a total. These can be written as dots, as totals or by using a symbol for each outcome.

Die 1

Die

2

1

2

3

4

5

6

1 2 3 4 5 6

Die 1

Die

2

1

2

3

4

5

6

1 2 3 4 5 6

2

3

4

5

6

7

3

4

5

6

7

8

4

5

6

7

8

9

5

6

7

8

9

10

6

7

8

9

10

11

7

8

9

10

11

12

Die

Coi

n

H

T

1 2 3 4 5 6

H1

T1

H2

T2

H3

T3

H4

T4

H5

T5

H6

T6

WorkEd ExamplE 11

A die is rolled twice.a Draw a lattice diagram to show all of the possible outcomes.b Find the probability of rolling a 2, then a 1.c Find the probability of getting a total of 7.


Think WriTE/draW

a The possible outcomes from each roll are 1, 2, 3, 4, 5, 6. Put these numbers on each axis.

a

1

2

3

4

5

6

1 2 3 4 5 6

b The probability of obtaining a 2 on the fi rst die and a 1 on the second is shown by one outcome only.There are 36 total possible outcomes.

b Pr(2, 1) = 136

c 1 The question asks for the probability of 2 numbers and the probability of a total of 7, so write totals on the diagram to show the possible outcomes of both events. There are 6 ways of getting a total of 7 from a total of 36 possible outcomes.{1 and 6, 2 and 5, 3 and 4, 4 and 3, 5 and 2, 6 and 1}

c

Die 1

Die

2

1

23

45

6

1 2 3 4 5 6

2

3

4

56

7

3

4

5

67

8

4

5

6

78

9

5

6

7

89

10

6

7

8

910

11

7

8

9

1011

12

2 Write the answer. Pr(total = 7) = 636

= 16

Exercise 11C Tree diagrams and lattice diagrams 1 WE9 A psychic powers test kit contains 10 blue, 10 red and 10 green cards, each without any

markings. In one particular test session, ‘Mental Mal’ selects a card, replaces it, and selects a card again.a Draw a tree diagram showing the possible colour outcomes at each stage.b Calculate the probability of Mal choosing: i two blue cards ii a red card, then a green card iii a green and a red card.

2 A coin is tossed together with a disc that is red on one side and white on its other side.a Show all possible outcomes on a tree diagram.b Calculate the probability that the coin lands Tails and the disc lands red.

3 Two letters from the word CAT are chosen.a Show all possible outcomes on a tree diagram.b Calculate the probability that the letter A is chosen fi rst and the letter T is chosen second.

4 Two coins are tossed.a Show all possible outcomes on a tree diagram.b Find the probability that one head and one tail turned up.

5 The two spinners shown are spun and the colour on which each stops is noted.Find the probability that the spinners land on:a red and greenb yellow and bluec yellow and green.

diGiTal doCdoc-9806Stirling’s formula

Spinner 2Spinner 1


6 mC A coin is tossed and a wheel that is coloured blue, white and yellow is spun.The probability of getting Tails and the colour yellow is:

a 56

B 16

C 45

d 12

E 34

7 A pentagonal solid whose faces are numbered 2, 4, 6, 8, 10 is rolled and a disc that is red on one side and blue on its other side is tossed. Draw a tree diagram and calculate the probability that a number greater than 4 is rolled and the colour showing uppermost on the disc is red.

8 WE 10 An integer from 2 to 3 inclusive is chosen from one hat and an integer from 4 to 6 inclusive is chosen from another hat.Draw a tree diagram showing the possible outcomes and determine the probability of selecting:a two even numbersb two odd numbersc two even numbers or two odd numbers.

9 Peter chooses to wear a jacket and tie from the available jackets and ties on his clothes rack, which is shown in the photo at right. Use the photo to draw a tree diagram showing the possible jacket and tie choices. Calculate the probability of choosing the darker brown jacket with the red and yellow tie.

10 Each of the smaller triangles formed by the intersection of the diagonals of a square is painted using either red, green or blue before covering each one with a low-sheen or full-gloss varnish. If the colour of each triangle is chosen at random, draw a suitable tree diagram and find the probability that the triangle is not coloured red or green and is covered with full-gloss varnish.

11 A coin is tossed three times.a Show all possible outcomes on a tree diagram.b Find the probability of getting Head, Tail, Tail.c What is the probability of getting at least two Tails?

12 Johnny wishes to try all combinations of a supercone ice-cream that has three scoops of different flavours chosen from chocolate, vanilla, strawberry, lime and banana. The middle scoop must be chocolate. If Johnny randomly chooses his supercone ice-cream, show all possible outcomes on a tree diagram.

13 Alan, Bjorn and Carl each toss a coin at the same time. Draw an appropriate tree diagram and use it to find the probability that Alan’s and Carl’s result will both be Tails.

14 A consonant is selected from each of the words MATHS IS FUN.a Show the possible outcomes on a tree diagram.b Find the probability that the letters H and S will appear in the selection.

15 Two coins are tossed and a die is rolled. One of the coins is double-headed. Find the probability that you get:a two Heads and an even numberb a Head, a Tail and an odd numberc a Head, a Tail and a number less than 4.

16 Find the probability of obtaining an odd number and at least one Tail when a die and two coins are tossed.

17 mC Three coins are tossed once. The probability that at least one coin shows Heads is:

a 34

B 38

C 23

d 78

E 14

Chocolate


18 Four coins are tossed.a Show all possible outcomes on a tree diagram.b Find the probability of obtaining Head, Tail, Head, Tail in that order.c Find the probability of obtaining two Heads and two Tails.d Find the probability of obtaining at least two consecutive Tails.

19 WE 11 a Draw a lattice diagram to show all possible outcomes when two dice are rolled.b Use the lattice diagram to find the probability that both the numbers appearing uppermost are odd

numbers.c Find the probability of getting a total of 9.

20 A die is rolled and a coin is tossed.a Draw a lattice diagram to show all of the possible outcomes.b Find the probability of obtaining a 3 and a Tail.c Find the probability of obtaining an even number and a Head.

21 Two dice are rolled. Find the probability:a of obtaining two 6sb of rolling a 3 and a 4c that the sum of the numbers appearing uppermost is

less than 10d that the first number is a 3 and the sum of the numbers

appearing uppermost is less than 8e of rolling two multiples of 2.

22 A die labelled with the letters T, O, M, A, T, O and a die numbered 3, 4, 5, 6, 7, 8 are rolled together. Determine the probability that the first die shows a vowel and the second die shows a number greater than 6.

23 A diner orders an entree, main course and dessert from a lunch menu that offers 3 different entrees, 2 different main courses and 2 different desserts. Show these choices on a tree diagram and find the probability that the diner orders a particular entree and main course.

11d The addition law of probabilitiesRecall from our review of set theory that:

n(A ∪ B) = n(A) + n(B) − n(A ∩ B) [1]

We also know that:

Pr(A ∪ B) = n(A ∪ B)

n(ξ ) [2]

Substituting [1] into [2], we get:

Pr(A ∪ B) = n(A) + n(B) − n(A ∩ B)

n(ξ )

= n(A)

n(ξ ) +

n(B)

n(ξ ) −

n(A ∩ B)

n(ξ )

So, Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B).Since we may equate ∪ with OR and ∩ with AND, we can say:

Pr(A or B) = Pr(A) + Pr(B) − Pr(A and B)

This is known as the Addition Law of Probabilities.

mutually exclusive eventsIf A ∩ B = ϕ, then A and B are mutually exclusive. That is, events A and B cannot happen at the same time. If A and B are mutually exclusive, Pr(A ∩ B) = 0, the Addition Law becomes:

Pr(A ∪ B) = Pr(A) + Pr(B)or

Pr(A or B) = Pr(A) + Pr(B)


WorkEd ExamplE 12

If A and B are events such that Pr(A) = 0.8, Pr(B) = 0.2 and Pr(A ∩ B) = 0.1, calculate Pr(A ∪ B).

Think WriTE

Substitute the values for Pr(A), Pr(B) and Pr(A ∩ B) in the Addition Law to fi nd Pr(A ∪ B).

Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B)Pr(A ∪ B) = 0.8 + 0.2 − 0.1Pr(A ∪ B) = 0.9

WorkEd ExamplE 13

If A and B are events such that Pr(A ∪ B) = 0.55, Pr(A) = 0.2 and Pr(B) = 0.45, calculate Pr(A ∩ B).

Think WriTE

1 Substitute the values for Pr(A ∪ B), Pr(A) and Pr(B).

Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B)0.55 = 0.2 + 0.45 − Pr(A ∩ B)

2 Rearrange the expression to fi nd Pr(A ∩ B). 0.55 = 0.65 − Pr(A ∩ B)Pr(A ∩ B) = 0.1

WorkEd ExamplE 14

If Pr(A ∩ B) = 0.2 and Pr(A ∪ B) = 0.9, calculate Pr(A) and Pr(B) if events A and B are equally likely to occur.

Think WriTE

1 Use the Addition Law. Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B)

2 If events A and B are equally likely to occur, then Pr(A) = Pr(B).

Let x represent Pr(A) and hence Pr(B).

3 Substitute the information into the Addition Law and solve.

0.9 = x + x − 0.20.9 = 2x − 0.21.1 = 2x

so x = 0.55Pr(A) = 0.55, Pr(B) = 0.55

WorkEd ExamplE 15

A box contains 16 marbles numbered 1, 2, 3, . . . , 16. One marble is randomly selected.Let A be the event ‘the marble selected is a prime number greater than 3’ and let B be the event ‘the marble selected is an odd number’.a Evaluate: i Pr(A) ii Pr(B) iii Pr(A ∩ B) iv Pr(A ∪ B).b Are A and B mutually exclusive events?

Think WriTE

a 1 Write down the elements of A, B, A ∩ B and A ∪ B.

a A = {5, 7, 11, 13} B = {1, 3, 5, 7, 9, 11, 13, 15} A ∩ B = {5, 7, 11, 13} A ∪ B = {1, 3, 5, 7, 9, 11, 13, 15}

2 What is n(ξ)? n (ξ) = 16



3 Calculate the probability of A, B, A ∩ B and A ∪ B.

i Pr(A) = 416

= 14

ii Pr(B) = 816

= 12

iii Pr(A ∩ B) = 416

= 14

iv Pr(A ∪ B) = 816

= 12

b Since Pr(A ∩ B) ≠ 0, it follows that A and B are not mutually exclusive.

b From iii above, we see that Pr(A ∩ B) = 14.

As Pr(A ∩ B) ≠ 0, A and B are not mutually exclusive.

Note: The Addition Law could also be used to determine any one of Pr(A ∪ B), Pr(A), Pr(B) or Pr(A ∩ B) when the other three quantities are known.

For example, to find Pr(A ∪ B) we have:

Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B)

= 416

+ 816

− 416

= 12

The Venn diagram below left may be adapted to show probabilities rather than outcomes and used to solve problems (below right).

ξ

A B

(A ∪ B)′ or

A ∩ BA ∩ B′ A′ ∩ B

A′ ∩ B′

ξ Pr(A ∪ B)' or Pr(A' ∪ B')

A BPr(A ∩ B)Pr(A ∩ B') Pr(A' ∩ B)

WorkEd ExamplE 16

An 8-sided die (numbered from 1 to 8) is rolled once. Find the probability that the number appearing uppermost is: a an even number b an even number or a multiple of 3.

Think WriTE

a 1 Let E = even number = {2, 4, 6, 8}.The probability of getting an even

number = n(E )

n(ξ ).

a Pr(E ) = n(E )

n(ξ )

2 n(E) = 4, n(ξ ) = 8 Pr(E) = 48

= 12

b 1 M = multiple of 3 = {3, 6}. The probability of an even number or a multiple of 3 = Pr(E ∪ M ).

b Pr(E ∪ M) = Pr(E) + Pr(M ) − Pr(E ∩ M )

2 Pr(E ) = 12, Pr(M ) = 2

8 = 1

4,

E ∩ M = {6} so Pr(E ∩ M ) = 18.

Pr(E ∪ M ) = 12 + 1

4 − 1

8

= 58


WorkEd ExamplE 17

If Pr(A) = 0.6, Pr(B) = 0.45 and Pr(A ∪ B) = 0.7, show this information on a Venn diagram and calculate Pr(A ∪ B)′.

Think WriTE/draW

1 Draw a 2-set Venn diagram with an overlapping region.

2 Calculate the probability of the overlap, Pr(A ∩ B), using the Addition Law.

Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B) 0.7 = 0.6 + 0.45 − Pr(A ∩ B) Pr(A ∩ B) = 1.05 − 0.7 = 0.35

3 Complete the Venn diagram using the available information.

ξA B

(A ∪ B)'

0.25 0.35 0.1

4 Calculate Pr(A ∪ B)′. Pr(A ∪ B) + Pr(A ∪ B)′ = 1 Pr(A ∪ B)′ = 1 − Pr(A ∪ B) = 1 − 0.7 = 0.3

Exercise 11d The addition law of probabilities 1 WE12 If Pr(A) = 0.4, Pr(B) = 0.5 and Pr(A ∩ B) = 0.2, what is Pr(A ∪ B)?

2 If Pr(A) = 0.65, Pr(B) = 0.25 and Pr(A ∩ B) = 0.22, what is Pr(A ∪ B)?

3 If A and B are mutually exclusive events and Pr(A) = 0.38, Pr(B) = 0.51, what is Pr(A ∪ B)?

4 WE13 If A and B are events such that Pr(A) = 0.4, Pr(B) = 0.5 and Pr(A ∪ B) = 0.6, calculate Pr(A ∩ B).

5 For events X, Y, if Pr(Y) = 0.44, Pr(X ∩ Y) = 0.16 and Pr(X ∪ Y) = 0.73, what is Pr(X)?

6 For events D and E, if Pr(D) = 0.76, Pr(D ∪ E) = 0.82 and Pr(D ∩ E) = 0.35, what is Pr(E)?

7 WE14 If Pr(A) = 2 × Pr(B), Pr(A ∩ B) = 0.23 and Pr(A ∪ B) = 0.94, determine the values of Pr(A) and Pr(B).

8 If Pr(A ∪ B) = 0.75, Pr(A) = 0.28 and Pr(B) = 0.47, what can be concluded about the relationship between A and B?

9 If Pr(A ∩ B) = Pr(A), what is the relationship between A and B?

10 WE15 A card is chosen at random from a pack of 52 playing cards. Let H be the event ‘choosing a heart’ and P be the event ‘choosing a picture card (J, Q, K)’.a Evaluate: i Pr(H) ii Pr(P) iii Pr(H ∩ P) iv Pr(H ∪ P).b Are H and P mutually exclusive events?

11 A box of chocolates contains 12 with hard centres and 8 with soft centres. One chocolate is chosen at random. Let H be the event ‘choosing a hard centre’ and S be the event ‘choosing a soft centre’.a Evaluate: i Pr(H) ii Pr(S) iii Pr(H ∩ S) iv Pr(H ∪ S).b Are H and S mutually exclusive events?


12 From a group of 100 people, 25 said they drink tea, 40 said they drink coffee and 15 said they drink both beverages. If one member of the group is randomly chosen, what is the probability that the person:a drinks only tea?b drinks neither tea nor coffee?c drinks tea and coffee?d drinks tea or coffee?

13 WE16 A box contains 20 marbles numbered 1 to 20. Find the probability of obtaining:a an even numberb a multiple of 3c a multiple of 2 or 3.

14 Find the probability of an odd number or a multiple of 4 appearing uppermost when a die is rolled.

15 Find the probability that a number divisible by 4 or 5 is drawn from a ‘lucky dip’ containing the first 50 natural numbers.

16 From a standard pack of 52 playing cards, one card is randomly drawn. State the probability that the card is:a a tenb a diamondc a king or a jackd a diamond, a spade or the ace of hearts.

17 A mixed bag of lollies contains 8 peppermint twists, 10 red jelly beans, 10 caramels, 18 chocolates, 4 peppermint twirls, 5 yellow jelly beans and 25 toffees. If Tara randomly selects one lolly, what is the probability that it is:a a peppermint or a jelly bean?b not a toffee, a caramel or a jelly bean?c a peppermint, given that the jelly beans are stuck together and cannot be selected?

18 Sarah is competing in a 400-metre race against 13 other runners. If each contestant has the same probability of winning, find the probability that Sarah:a wins the raceb comes first or secondc finishes in the top fourd does not qualify for the final 5.

19 A moneybox containing eight $1 coins, five $2 coins, nine 50c coins and two 20c coins is shaken and one coin falls out. Assuming that each coin is equally likely to fall out, calculate the probability that the coin’s value is:a between 10c and $2 (not including 10c or $2)b not 50cc $1 or $2d less than $1.

20 Inside a dresser drawer are 4 ties, 10 socks, 4 handkerchiefs and 2 towels. If Tony randomly takes out one item, find the probability that it is:a something to be wornb not a towel and not a sockc either a sock or not a sockd either a towel or not a handkerchief.

21 A patron in a restaurant is presented with a fruit platter consisting of 6 whole apples, 8 slices of orange, 5 sliced pear pieces, 11 whole strawberries, 6 whole plums and 4 sliced apricot halves. The waiter accidentally trips and a piece of fruit falls off the platter. Assuming that each piece of fruit was equally likely to fall, state the probability that the fallen fruit is:a not a plum and not an apricotb not slicedc sliced or is not a strawberryd either a pear or an orange that has not been sliced.

22 The games Alotto, Blotto and Clotto involve guessing a number from 1 to 100 inclusive. To win Alotto the number guessed must be a multiple of 3. To win Blotto the number must be a multiple of 5 or a multiple of 8. To win Clotto the guessed number is to be between 10 and 20 or greater than 77.

Decide which game is easier to win.


23 WE17 If Pr(A) = 0.3, Pr(B) = 0.4 and Pr(A ∪ B) = 0.65, show this information on a Venn diagram and find Pr(A′ ∪ B).

24 Of 20 people interviewed, 7 stated that they use both a tram and a train to get to work, and 2 said they drive their own car. No other form of transport or combination of transport is used. If 5 people travel only by train, find the probability that a person selected at random travels by tram only.

25 The unusual dartboard shown below consists of 10 concentric circles, with 1024 points given for a dart landing within the first (smallest) circle, 512 points for a hit within the area bounded by the first and second circle, 256 points if the dart lands within the area bounded by the second and third circles, and so on. The area bounded by any two consecutive circles is the same.

Area 11024 points

Area 2512 points

Area 3256 points

Area 4128 points

a Find the probability that a dart randomly hitting the board will score: i 64 ii a multiple of 128 iii a number from 16 to 256 inclusive iv a number from 17 to 1023 inclusive or a number less than 256.b Why is it necessary to state that the areas bound by any two consecutive circles are

the same?

11E karnaugh maps and probability tablesWe have seen how Venn diagrams provide a visual representation of sets and probabilities. Another effective approach is to display the information by means of a Karnaugh map. Consider a Venn diagram for two sets A and B.

ξA B

A ∩ B' A ∩ B A' ∩ B

A' ∩ B'

Notice that the Venn diagram consists of four mutually exclusive regions, A ∩ B′, A ∩ B, A′ ∩ B and A′ ∩ B′. These four subsets of ξ can be presented as a Karnaugh map.

Column 1 Column 2 Column 3

B B′Row 1 A A ∩ B A ∩ B′Row 2 A′ A′ ∩ B A′ ∩ B′Row 3

diGiTal doCdoc-9807

WorkSHEET 11.2


Comparing the table entries with the Venn diagram provides equality relationships across rows and down columns. That is, in terms of regions we can see that for column 1, (A ∩ B) ∪ (A′ ∩ B) = B, and for column 2, (A ∩ B′) ∪ (A′ ∩ B′) = B′.

Similarly, for row 1, (A ∩ B) ∪ (A ∩ B′) = A, and for row 2, (A′ ∩ B) ∪ (A′ ∩ B′) = A′. The third row and column can be used to check the sum totals of each row and column. This type of verifi cation can be useful in practical problems.

The probability tableWe can present a Karnaugh map in terms of the probability of each of the four subsets A ∩ B′, A ∩ B, A′ ∩ B and A′ ∩ B′ of ξ.

B B′

A Pr(A ∩ B) Pr(A ∩ B′) Pr(A)

A′ Pr(A′ ∩ B) Pr(A′ ∩ B′) Pr(A′)

Pr(B) Pr(B′) 1

Note the value of 1 at the bottom right of the table. This is the sum of the probabilities across the last row and the sum of the probabilities down the last column.

That is, Pr(B) + Pr(B′) = 1 and Pr(A) + Pr(A′) = 1.Consider the following example. A survey of 1000 taxi drivers revealed that 450 of them drive

Falcons and 500 drive Commodores. It was also found that 350 taxi drivers have occasion to use both types of car. This information can be represented as a Venn diagram, a Karnaugh map or a probability table.

ξF C

150350100

400

C C′

Row 1 F 350 100 450

Row 2 F′ 150 400 550

Row 3 500 500 1000

C C′

F 0.35 0.1 0.45

F′ 0.15 0.4 0.55

0.5 0.5 1.0

Venn diagram Karnaugh map Probability table

The Karnaugh map provides the following information:1. 350 drivers drive both a Falcon and a Commodore (row 1, column 1: F ∩ C ).2. 100 drivers drive only a Falcon (row 1, column 2: F ∩ C ′).3. The total number of Falcon drivers is 450 (350 + 100).4. 150 drivers drive only a Commodore (row 2, column 1: F ′ ∩ C ).5. 400 drivers do not drive either a Falcon or a Commodore (row 2, column 2: F ′ ∩ C ′).6. There are 500 Commodore drivers altogether (350 + 150).7. There are 1000 drivers in total (row 3, column 3).

WorkEd ExamplE 18

Complete the probability table shown below and represent the information as a Venn diagram.

Column 1 Column 2 Column 3

B B′

Row 1 A 0.3

Row 2 A′ 0.25

Row 3 0.65 1



Think WriTE/draW

1 Find the value for row 2, column 3 and for row 3, column 2.

B B′A 0.3

A′ 0.25 0.7

0.65 0.35 1

2 Find the value for row 2, column 1 and for row 1, column 2.

B B′A 0.1 0.3

A′ 0.45 0.25 0.7

0.65 0.35 1

3 Find the value for row 1, column 1. B B′A 0.2 0.1 0.3

A′ 0.45 0.25 0.7

0.65 0.35 1

4 Represent the information as a Venn diagram. ξA B

0.450.20.1

0.25

WorkEd ExamplE 19

Complete a probability table, given that Pr(A′ ∩ B) = 0.24, Pr(A) = 0.32 and Pr(B) = 0.35.

Think WriTE

1 Place the known information in the appropriate cells of the probability table.

B B′A 0.32

A′ 0.24

0.35 1

2 Build up the table using the given information and the fact that the probability totals 1.

B B′A 0.11 0.32

A′ 0.24 0.68

0.35 0.65 1

B B′A 0.11 0.21 0.32

A′ 0.24 0.44 0.68

0.35 0.65 1

WorkEd ExamplE 20

A group was surveyed in relation to their drinking of tea and coffee. From the results it was established that if a member of the group is randomly chosen, the probability that that member drinks tea is 0.5, the probability that they drink coffee is 0.6, and the probability that they drink neither tea nor coffee is 0.1.a Use the information to complete a probability table.b Calculate the probability that a randomly selected person of the group: i drinks tea but not coffee ii drinks tea and coffee.


Think WriTE

a 1 Let T and C be the set of people who drink tea and coffee respectively. Place the given information in the table.

a C C′T 0.5

T′ 0.1

0.6 1

C C′T 0.2 0.3 0.5

T′ 0.4 0.1 0.5

0.6 0.4 1

2 Build up the table as shown.

C C′T 0.5

T′ 0.1 0.5

0.6 0.4 1

C C′T 0.3 0.5

T′ 0.4 0.1 0.5

0.6 0.4 1

b 3 Use the appropriate probability from the table.

b i Pr(T ∩ C′ ) = 0.3ii Pr(T ∩ C ) = 0.2

Exercise 11E karnaugh maps and probability tables1 WE 18 Complete each Karnaugh map and represent the information as a Venn diagram.

a B B′

A 17 25

A′ 13

15

b B B′

A 33

A′ 27 72

114

c B B′A 0.3 0.57

A′ 0.4

d B B′A 0.03

A′ 0.22 0.36

2 mC Decide which of the following statements is true.

a U = 0.15 B V + W = 0.42 C X + Y = 0.55d V − X = W − 0.58 E U + Z = W − Z

3 Complete a Karnaugh map given that n(A ∩ B) = 87, n(A′ ∩ B) = 13, n(A ∩ B′) = 63 and n(ξ ) = 218.

4 Complete a Karnaugh map given that n(A ∩ B) = 35, n(A ∩ B′) = 29, n(A′ ∩ B′) = 44 and n(A′ ∩ B) = 56.

5 Draw a Karnaugh map representing each Venn diagram.a ξ

A B

0.3

0.11

0.450.14

b ξA B

0.120.610.27

B B′A 0.31 Y 0.75

A′ X U Z

V 0.58 W


c ξA B

4610

d A B

18

5

715

ξA B

18

5

715

6 Determine the probability values and complete a probability table using the given information.a ξ = {letters of the alphabet from a to k}, A = {a, b, c, d, e, f, g}, B = {e, f, g, h}b ξ = {fi rst 20 natural numbers}, A = {natural numbers less than 11}, B = {natural numbers from 8

to 15 inclusive}

7 mC If A = {2, 7, 8, 10}, B = {3, 5, 7, 9, 10} and ξ = {1, 2, . . . , 10}, then A ∩ B′ will contain the set:a {3, 5, 7} B {1, 4} C {2, 6, 9}d {6, 7, 10} E {2, 8}

8 A survey of students revealed that 30 of them like football, 26 like soccer, 6 like both sports and 10 prefer a sport other than football or soccer. Represent this information as a:a Venn diagram b Karnaugh map.

9 Complete a probability table for the information in question 8.

10 mC Of a group of 200 people, 48% drink coffee (C ) each day and 39% drink tea (T ). If 38% of the people do not drink tea or coffee, the probability table is:a T T′

C 0.25 0.23

C′ 0.14 0.38

B T T′C 0.23 0.14

C′ 0.38 0.25

C T T′C 0.25 0.35

C′ 0.48 0.38

d T T′C 0.39 0.38

C′ 0.48 0.23

E T T′C 0.38 0.10

C′ 0.01 0.51

11 WE 19 Complete a probability table, given:

a Pr(A ∩ B) = 0.3, Pr(A′ ∩ B′) = 0.2 and Pr(A) = 0.6b Pr(B ∩ A′) = 0.7, Pr(B) = 0.8 and Pr(B′ ∩ A) = 0.1c Pr(A ∩ B) = 0.5, Pr(A′ ∩ B′) = 0.1 and Pr(B′) = 0.4d Pr(A′ ∩ B) = 1

4, Pr(A ∩ B′) = 1

4 and Pr(B′) = 3

4.

12 Two hundred and eighty children were asked to indicate their preference for ice-cream flavours. It was found that 160 of the children like chocolate flavour, 145 like strawberry and 50 like both flavours. Use this information to complete a Karnaugh map.

13 WE20 An examination of 250 people showed that of those in the group who are less than or equal to 20 years of age, 80 wear glasses and 55 do not. Also, 110 people over 20 years of age must wear glasses.a Represent the information as a probability table.b Calculate the probability that a randomly selected person of the group: i does not wear glasses and is over 20 years of age ii is 20 years of age or younger.

14 For the probability table shown, A is the event ‘no more than 15 years of age’ and B is the event ‘smoker’.a Complete the probability table.b What is the probability that: i a person older than 15 years of age does not smoke? ii a person is a smoker and is older than 15 years of age? iii the person is a smoker over the age of 15 or is a non-

smoker less than or equal to 15 years of age?

B B′A 0.08

A′ 0.6

0.67


15 A survey of a group of business people indicates that 42% of those surveyed read the Daily Times newspaper only each day and 18% read both the Daily Times and the Bugle. Additionally, 12% of those questioned stated that they do not read either of these newspapers.a Show this information as a Karnaugh map.b What is the probability that a randomly selected

member of the group:i does not read the Daily Times?ii reads the Bugle only?iii does not read the Bugle or does not read either

newspaper?c If the group consists of 150 business people, determine

how many members read at least one newspaper.

16 A lucky dip box contains 80 marble tokens that can be exchanged for prizes. Some of the marbles have a red stripe, some have a blue stripe, some have both a red and a blue stripe, and some marbles have no stripes at all. It is known that 25% of the marbles have a red stripe on them, 20% of them have a blue stripe and 65% have no stripe.a Present the information as a Karnaugh map.b What is the probability of choosing a marble that has a red stripe only?c Find the probability of choosing a marble that has a red and a blue stripe or no stripe.

11F Conditional probabilityErin thinks of a number from 1 to 10 (say 8) and asks Peter to guess what it is. The probability that Peter makes a correct guess on his first try is 1

10. If, however, Erin first tells Peter that the number is greater

than 7, his chances are better because he now knows that the number must be one of the numbers 8, 9 or 10. His probability of success is now 1

3.

This problem may be stated as: What is the probability of Peter choosing the right number from 1 to 10, given that the number is greater than 7?

This is an example of conditional probability, where the probability of an event is conditional on (that is, it depends on) another event occurring first. The effect in this case is to reduce the event space and thus increase the probability of the desired outcome.

For two events A and B, the conditional probability of event A given that event B occurs is denoted by Pr(A | B) and is given by:

Pr(A | B) = Pr(A ∩ B)

Pr(B), Pr (B) ≠ 0

Event B is sometimes called the reduced event space.For the example above, if we let B be the event ‘numbers greater than 7’ and A be the event

‘Erin’s secret number’, then we may write:

Pr(A | B) = Pr(A ∩ B)

Pr(B)

= 110310

= ( 110

÷ 310

)

= 13

The reduced event space can be illustrated by the Venn diagram below.

ξA

B

8 9 10

1234

567


WorkEd ExamplE 21

If Pr(A ∩ B) = 0.8 and Pr(B) = 0.9, find Pr(A | B).

Think WriTE

Substitute the values given into the expression for conditional probability.

Pr(A | B) = Pr(A ∩ B)

Pr(B)

= 0.80.9

= 89

WorkEd ExamplE 22

If Pr(A) = 0.3, Pr(B) = 0.5 and Pr(A ∪ B) = 0.6, calculate:a Pr(A ∩ B) b Pr(A | B).

Think WriTE

a Use the Addition Law for probabilities to find Pr(A ∩ B).

a Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B) 0.6 = 0.3 + 0.5 − Pr(A ∩ B)so Pr(A ∩ B) = 0.2

b Use the formula for conditional probability to find Pr(A | B).

b Pr(A | B) = Pr(A ∩ B)

Pr(B)

= 0.20.5

= 25

WorkEd ExamplE 23

Of a group of 50 Year 11 students, 32 study Art and 30 study Graphics. Each student studies at least one of these subjects.a How many students study both?b Illustrate the information as a Venn diagram.c What is the probability that a randomly selected student studies Art only?d Find the probability that a student selected at random from the group studies Graphics, given

that the student studies Art.

Think WriTE/draW

a 1 Define relevant events. a Let A = students who study Art G = students who study GraphicsA ∩ G = students who study both

2 Find the number who study both subjects using set theory.

n(A ∪ G) = n(A) + n(G) − n(A ∩ G) 50 = 32 + 30 − n(A ∩ G) 50 = 62 − n(A ∩ G)n(A ∩ G) = 12

b Show all the information on a Venn diagram. b ξArt Graphics

20 12 18

c The Venn diagram reveals that 20 of the 50 students study Art only. Calculate the probability.

c Pr(Art only) = 2050

= 25


d Use the conditional probability formula to fi nd the probability that a student studies Graphics, given that the student studies Art.

d Pr(G | A) = Pr(G ∩ A)

Pr(A)

= 1250

÷ 3250

= 1232

= 38

WorkEd ExamplE 24

Seated in a Ford Falcon are 4 males and 2 females. Seated in a Holden Commodore are 2 males and 1 female. One of the cars is randomly stopped by the police and one person from the vehicle is randomly selected.

Draw a tree diagram to illustrate the situation and calculate the probability that:a the person selected by the police is femaleb if a female is selected by the police, she was sitting in the Ford.

Think WriTE/draW

a 1 Calculate the probabilities.

a Pr(Ford) = 12 Pr(Holden) = 1

2

Pr(male from Ford) = 46 Pr(male from Holden) = 2

3

= 23

Pr(female from Ford) = 26 Pr(female from Holden) = 1

3

= 13

2 Draw the tree diagram.

male Ford and male

female Ford and female

male Holden and male

female Holden and female

Holden

Ford

Car Person

1–2

1–2

2–3

1–3

2–3

1–3

3 Use the tree diagram to work out the probability that the person is female. Consider all the ways a female may be selected.

Pr(female selected) = Pr(Ford and female or Holden and female)= Pr(Ford and female) + Pr(Holden and female)

= 12 × 1

3 + 1

2 × 1

3

= 13

b Use the tree diagram and the formula for conditional probability.

b Pr(person is from the Ford | female is selected)

=

=×

=

Pr(Ford and female)

Pr(female)12

13

13

12



Once the tree diagram was drawn, the calculation for part a in worked example 24 was quite intuitive. In order to calculate the probability of a female being selected, the occupants of both cars needed to be considered. In fact, there is a rule of probability that formalises the calculation performed in part a above. The law is known as the Law of Total Probability, and it states:

Pr(A) = Pr(A | B)Pr(B) + Pr(A | B′)Pr(B′)

To calculate the answer to part a of worked example 24, let Pr(A) = Pr(female) and Pr(B) = Pr(Ford).

Pr(female) = Pr(female given the car is a Ford)Pr(Ford)+ Pr(female given the car is a Holden)Pr(Holden)

= 13 × 1

2 + 1

3 × 1

2

= 13

Note that the Law of Total Probability simplifi es to give the rule used in part a of worked example 24:

Pr(A) = Pr(A | B)Pr(B) + Pr(A | B′)Pr(B′)Pr(A) = Pr(A ∩ B) + Pr(A ∩ B′)

Exercise 11F Conditional probability 1 WE21 If Pr(A) = 0.8, Pr(B) = 0.5 and Pr(A ∩ B) = 0.4, find:

a Pr(A | B) b Pr(B | A).

2 If Pr(A) = 0.65, Pr(B) = 0.75 and Pr(A ∩ B) = 0.45, find:a Pr(A | B) b Pr(B | A).

3 If Pr(A ∩ B) = 0.4 and Pr(A) = 0.5, find Pr(B | A).

4 If Pr(A ∩ B) = 0.25 and Pr(B) = 0.6, find Pr(A | B).

5 If Pr(B | A) = 0.32 and Pr(A) = 0.45, find Pr(A ∩ B).

6 If Pr(A | B) = 0.21 and Pr(B) = 0.8, what is Pr(A ∩ B)?

7 Calculate Pr(A) if Pr(B | A) = 0.75 and Pr(A ∩ B) = 0.5.

8 Calculate Pr(B) if Pr(A | B) = 0.96 and Pr(A ∩ B) = 0.8.

9 WE22 If Pr(A) = 0.7, Pr(B) = 0.5 and Pr(A ∪ B) = 0.9, calculate:a Pr(A ∩ B)b Pr(B | A).

10 mC If Pr(B | A) = 0.8 and Pr(A ∩ B) = 0.6, then Pr(A) is:

a 45

B 35

C 14

d 34

E 23

11 mC If Pr(A) = 0.9 and 2 × Pr(A ∩ B) = Pr(A), then Pr(B | A) is:

a 12

B 59

C 25

d 19

E 49

12 Show that if Pr(A ∩ B) = Pr(A) × Pr(B), then Pr(B | A) = Pr(B).

13 If Pr(A) = 0.23, Pr(B) = 0.27 and Pr(A ∪ B) = 0.3, find:a Pr(A ∩ B) b Pr(A | B).

14 If Pr(A) = 0.45, Pr(B) = 0.52 and Pr(A ∪ B) = 0.67:a fi nd Pr(A ∩ B)b fi nd Pr(B | A)c represent the information as a Venn diagram.

15 A box contains marbles numbered 1, 2, 3, . . . 50. One marble is randomly taken out of the box. What is the probability that it is:a a multiple of 3, given that it is less than 21?b between 11 to 39 inclusive, given that it is greater than 20?

diGiTal doCdoc-9808

SkillSHEET 11.1Conditional probability


16 mC A group of 80 females consists of 54 dancers and 35 singers. Each member of the group is either a dancer or a singer or both. The probability that a randomly selected member of the group is a singer given that she is a dancer is:a 0.17 B 0.44C 0.68 d 0.11E 0.78

17 WE23 A group of 60 adventurers comprises 30 mountain climbers and 45 scuba divers. If each adventurer does at least one of these activities:a How many adventurers are both climbers and

divers?b Illustrate the information on a Venn diagram.c What is the probability that a randomly

selected group member is a scuba diver only?d Find the probability that an adventurer

randomly selected from the group is a scuba diver, given that the adventurer is a mountain climber.

18 Of 200 families surveyed, 85% have a TV and 70% possess a CD player. Assuming each family has at least one of these items, what is the probability that one family randomly selected has a TV, given that they also own a CD player?

19 During the Christmas holidays 42 students from a group of 85 VCE students found vacation employment while 73 students went away on holidays. Assuming that every student had at least a job or went on a holiday, what is the probability that a randomly selected student worked throughout the holidays (that is, did not go away on holidays), given that he/she had a job?

20 WE24 The probability that a machine in a chocolate factory does not coat a SNAP chocolate bar adequately, therefore producing a defective product, is 0.08. The probability that it does not coat a BUZZ chocolate bar adequately is 0.11. On any day the machine coats 250 SNAP bars and 500 BUZZ bars. A chocolate bar is chosen at random from the production line. Draw a tree diagram to illustrate the situation and find the probability that the chocolate bar chosen at random is:a a BUZZ chocolate barb a SNAP chocolate bar and is defectivec defective, given that a SNAP bar is chosen.

21 The staff at Happy Secondary College is made up of 43 females and 29 males. Also, 22% of the females are under 40 years old, and 19% of the males are under 40. If a staff member is selected at random, what is the probability that:a a male is selected?b a male 40 years or over is selected?c a female under the age of 40 is selected?d a person under 40 years of age is selected?e the person is a female given that the person selected is under 40 years of age?

22 Two letters are randomly picked from the word INFINITESIMAL. If a letter can be used more than once, calculate the probability that both letters selected are vowels, given that the first letter is a vowel.

11G Transition matrices and markov chainsintroductionIn chapter 7 we saw many uses for matrices, from displaying information in an organised manner to solving simultaneous equations or representing transformations. Matrices are also very useful in certain conditional probability problems.

inTEraCTiViTYint-0270Transition matrices


ExampleA jar contains six red balls and four green balls. A ball is selected at random and not replaced. A second ball is then selected. Find the probability that the second ball is a red ball.

SolutionThe tree diagram illustrates the situation.

We may express the answer (where R2 is the event ‘selecting a red ball on the second selection’) in the form:

Pr(R2) = Pr(R2 | R1)Pr(R1) + Pr(R2 | G1 )Pr(G1)

= 59 × 6

10 + 6

9 × 4

10

= 5490

= 35

This is an example of the law of total probability, which may be stated as:

Pr(A) = Pr(A | B ) × Pr(B ) + Pr(A | B′) × Pr(B′)Applying the law to the complement of event A gives:

Pr(A′ ) = Pr(A′ | B) × Pr(B) + Pr(A′ | B′ ) × Pr(B′ )These two equations may be written in matrix form:

′

=| | ′

′ | ′ | ′

×′

A

A

A B A B

A B A B

B

B

Pr( )

Pr( )

Pr( ) Pr( )

Pr( ) Pr( )

Pr( )

Pr( )

The example can now be written in matrix form, where:A = selecting a red ball on the second selectionA′ = selecting a green ball on the second selectionB = selecting a red ball on the first selectionB′ = selecting a green ball on the first selection.

′

=

×′

=

×

=

A

A

B

B

Pr( )

Pr( )

59

69

49

39

Pr( )

Pr( )

59

69

49

39

610410

3525

The first element in the final column matrix is the same as the answer we obtained in the example shown. The second element is the probability of selecting a green ball on the second selection.

Note that the columns of the matrix each add to one. This is equivalent to one of the properties of probability, Pr(A) + Pr(A′ ) = 1. In this case it is actually Pr(A | B) + Pr(A′ | B) = 1 for the first column and Pr(A | B′ ) + Pr(A′ | B′ ) = 1 for the second column.

The matrix of conditional probabilities is called a transition matrix, usually denoted T.

T =

′

′| | ′

′ | ′ | ′

B B

AA

A B A B

A B A B

Pr( ) Pr( )

Pr( ) Pr( )

The preceding example can be thought of as a transition from an initial state (selection of the first ball, B or B′ being ‘red ball’ or ‘green ball’ respectively) to the next state (selection of the second ball, A or A′, that is ‘red ball’ or ‘green ball’).

The column matrices ′

AA

Pr( )Pr( )

and ′

BB

Pr( )Pr( )

are called state matrices, where ′

BB

Pr( )Pr( )

is the

initial state, which we might label as S0, and ′

AA

Pr( )Pr( )

is the next state, S1. The matrix equation becomes S1 = T × S0.

6—10

4—10

6–9

5–9

4–9

3–9

R1

G1

R2

G2

R2

G2


WorkEd ExamplE 25

Consider a simple model of the behaviour of a football team. If it wins a game, then the probability that it wins the next game is 0.8. If it loses, then the probability that it wins the next game is only 0.5. Write the transition matrix that represents the transition from one game to the next.

Think WriTE

1 Set up a table showing the information given.B (Wins fi rst game)

B′ (Loses fi rst game)

A (Winssecond game)

(Pr(A | B))0.8

(Pr(A | B′))0.5

A′ (Losessecond game)

(Pr(A′ | B))?

(Pr(A′ | B′))?

2 Complete the table using the knowledge that the columns must add to one.

B (Wins fi rst game)

B′ (Loses fi rst game)

A (Winssecond game)

(Pr(A | B) )0.8

(Pr(A | B′) )0.5

A′ (Losessecond game)

(Pr(A′ | B) )0.2

(Pr(A′ | B′))0.5

3 Write the answer.=

T 0.8 0.50.2 0.5

The matrix equation S1 = T × S0 describes the transition from state S0 (the fi rst game) to state S1 (the second game). If the conditional probabilities remain the same, then a similar equation will express the transition from any particular state to the next state.

In general, Sn + 1 = T × Sn may be used to determine any state from the previous one.

WorkEd ExamplE 26

Suppose the conditional probabilities expressed in worked example 25 remain constant throughout the football season. Also, suppose the team wins the fi rst game of the season. Express the problem in matrix form and fi nd the probability that the team loses the third game of the season.

Think WriTE

1 Write down the transition matrix. This is the same as in the previous worked example.

T =

0.8 0.50.2 0.5

2 Write down a suitable initial state matrix that shows the probabilities of the fi rst game. The fi rst element will be the probability that the team wins. Since we know it has won, this probability must be 1. The second element must be 0 as the column sums to 1.

S0 =

01



3 Enter the transition matrix, T, into a CAS calculator.

→ t0.8 0.50.2 0.5

4 Enter the initial state matrix, S0, into the CAS calculator.

→ s

10 0

5 Perform matrix multiplication to calculate S1 = T × S0.

t × s0 → s1

6 Record the result.

0.80.2

7 Perform matrix multiplication to calculate S2 = T × S1.

t × s1 → s2


0.740.26

9 The answer will be the second element of the state matrix as we want the probability of a loss.

The probability of losing game 3 is 0.26.

This worked example shows the power of using matrices. If the conditional probabilities remain constant and the outcomes of any particular state depend only on the previous state, then we have a (two-state) Markov Process. We can easily calculate the probabilities associated with any of the later states without determining all the intermediate ones.

As S1 = T × S0 and S2 = T × S1,S2 = T × T × S0 = T2 × S0

Continuing in this fashion, we see that:

S3 = T × S2 = T × (T × S1) = T × T × T × S0 = T3 × S0

and, in general, Sn = T n × S0.

WorkEd ExamplE 27

Using the data of worked example 25, fi nd the probability that the team wins the fi fth game of the season, assuming it loses the fi rst game.

Think WriTE

1 Write down the transition matrix.T =

0.8 0.50.2 0.5

2 Write down a suitable initial state matrix. We know that the team loses the fi rst game, so the second element must have a probability of 1.

S0 =

01

3 Identify which state matrix is required. Since S0 corresponds to game 1, game 5 must correspond to state matrix S4.


→ t0.8 0.50.2 0.5


→ s

01 0


6 Calculate the probabilities of the outcome of game 5 using the rule S4 = T 4 × S0.

t 4 × s0


0.70850.2915

8 Write down the answer. As we are interested in the probability of winning the fi fth game, we look at the fi rst element of the matrix.

The probability of winning the fi fth game, given that the team lost the fi rst game, is 0.7085.

alternative forms of the state matrixIt is not necessary to express a state matrix in terms of probabilities. We can use percentages or even raw numbers, as the following example illustrates.

WorkEd ExamplE 28

Suppose there are 800 people in a town who watch the two current affairs shows on television: Breakdown and News Roundup. Assume they all watch just one of the shows every week night. Also assume that if a person were to watch Breakdown on a particular night, then there is a probability of 0.35 that she will watch Breakdown the next night. If she were to watch News Roundup on a particular night, there is a probability of 0.45 that she will watch Breakdown the next night. These probabilities remain constant. If 300 people watch Breakdown on Tuesday night of a particular week, how many will watch each show on the next Thursday night?

Think WriTE

1 Set up a table that shows the information given.

Watches Breakdown on fi rst night

Watches News Roundup on fi rst night

Watches Breakdown on next night

0.35 0.45

Watches News Roundup on next night

2 Complete the table using the knowledge that the columns must add to one.

Watches Breakdown on fi rst night

Watches News Roundup on fi rst night

Watches Breakdown on next night

0.35 0.45

Watches News Roundup on next night

0.65 0.55

3 Write the transition matrix.=

T 0.35 0.450.65 0.55

4 The initial state matrix must show how many watch each show on the Tuesday night. If 300 watch Breakdown, then 500 (800 − 300) must watch News Roundup.

S0 =

300500


5 Idenfi ty which state matrix is required. Since S0 corresponds to Tuesday night, then Thursday night must correspond to state matrix S2.


→ t0.35 0.450.65 0.55


→ s

300500 0

8 Calculate the number of people who watch each show on Thursday night. Use S2 = T 2 × S0.

t 2 × s0


327473

10 Write the answer, remembering to round off to the nearest whole number if necessary.

327 watch Breakdown and 473 watch News Roundup on Thursday night.

Exercise 11G Transition matrices and markov chains 1 WE25 Consider a simple model of the behaviour of a netball team. If it wins a game, then

the probability that it wins the next game is 0.75. If it loses, then the probability that it wins the next game is only 0.5. Write down the transition matrix that represents the transition from one game to the next.

2 WE26 Suppose the conditional probabilities expressed in question 1 remain constant throughout the netball season. Also, suppose the team wins the first game of the season. Express the problem in matrix form and find the probability that the team loses the third game of the season.

3 WE27 For a transition matrix T =

0.6 0.70.4 0.3

and an initial state matrix S0 =

3070

, calculate:

a S1 b S2 c S3.

4 For a transition matrix T =

0.35 0.50.65 0.5

and an initial state matrix S0 =

800200

, calculate, giving

answers to the nearest whole number:a S1 b S2 c S4.


0.6 0.250.4 0.75

, calculate T 3. Use the result to calculate S3 if the initial

state matrix S0 =

10

.


0.85 0.50.15 0.5

, calculate T 4. Use the result to calculate S4 if the initial

state matrix S0 =

10

.

7 WE28 A school canteen offers vegetable or tomato soup in a cup each day. It is noticed that 30% of students who have vegetable soup on a given day select tomato soup the next day, and 60% who have tomato soup choose vegetable soup on the next day. There are 500 students who use the canteen each day, and they all have vegetable or tomato soup, but not both. On a particular Monday, 200 have vegetable soup and 300 have tomato soup.a Set up a transition matrix for this situation.b Write down the initial state matrix.


c How many students will have vegetable soup and how many will have tomato soup on the Friday of that week? (Answers must be given to the nearest whole number.)

d How many will have each type of soup on the following Friday? (Note that the canteen does not open on the weekend.)

8 Frank Viccuci is the goalkeeper for his soccer team. Assume that he always dives sideways when defending penalty kicks. If he dives to the right on a given occasion, then the probability that he dives to the right on the next occasion is 0.4. The transition matrix for the probabilities of Frank diving to

either side, given the side he dived to on the previous occasion, is

0.4 0.80.6 0.2

.

a If Frank dives to the left on a penalty he defends in today’s game, what is the probability that he will dive to the right for the next penalty?

b Suppose there is a penalty shoot-out that consists of each team taking five penalty shots at goal. The player taking the fifth penalty kicks to the right. If Frank had dived to the left on the first penalty, what is the probability that he dives to the right on the fifth penalty? Give answers accurate to 2 decimal places.

9 Assume that the probability of a particular football team winning its next game is 0.75 if it won its previous game and 0.55 if it lost its previous game. a If the team was successful in the opening game of the season, calculate the probability that it

will win:i the second game of the seasonii the fifth game of the seasoniii the tenth game of the seasoniv the twelfth game of the season.

Give answers accurate to 4 decimal places.b Repeat the calculations to find the corresponding probabilities if the team lost its opening game. c Can you predict, without further calculation, what is likely to happen in the last game of the

season? (Assume the season lasts for twenty-four games.)

10 Tasha likes to vary her gym routines in a special way. If she does a cycling class one day, the probability that she will do a Pilates session the next day is 0.6, and if she does a Pilates session one day, the probability she does cycling next day is 0.75. Assume that Tasha goes to the gym every day and does only cycling or Pilates.a Write down the transition matrix for this situation.b If Tasha cycles on Saturday, what is the probability that she also cycles the next Tuesday? c What is the probability that she does Pilates on that Tuesday?d What is the probability that she does Pilates on the next Tuesday?

11h independent eventsTwo events A and B are independent if each event has no effect on the likelihood of the other.

Consider two independent events A and B, where event A follows event B. If the probability of event A is unaffected by event B, then we can say that the probability of A, given B has happened, is the same as the probability of A (whether or not B has happened), or, using symbols:

Pr(A | B) = Pr(A) [1]

But | = ∩A B

A B

BPr( )

Pr ( )

Pr ( ) using the conditional probability formula. Rearranging the above equation

we have:

Pr(A ∩ B) = Pr(B) Pr(A | B) [2]

Note: Equation [2] has wide application in probability. It may be extended and interpreted as: ‘When calculating the probability of a chain of events, you may simply multiply by the probability of the next event, as long as the effect of previous events is taken into account’.

Substituting [1] into [2] we have:

Pr(A ∩ B) = Pr(A) Pr(B)


1. Pr(A ∩ B) means the probability of events A and B occurring.2. If Pr(A ∩ B) = Pr(A) × Pr(B), then the events A and B are independent.

One of the 12 outcomes possible when a coin and a die are simultaneously tossed is a Head for the coin and a 5 on the die. The number 5 obtained with the die does not come about because the coin comes up a Head, and getting a Head with the coin is not a result of the number 5 appearing uppermost on the die.

We can verify the expression given for independent events by looking further at the example of the coin and die.

What is the probability of getting a Tail and a number from 3 to 4 inclusive?The event space is ξ = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}, with n(ξ ) = 12.Let A be the event ‘getting a Tail’ and B be the event ‘getting a number from 3 to 4 inclusive’.

Then A ∩ B = {T3, T4}, n(ξ) = 12, so Pr(A ∩ B) = n(A ∩ B)

n(ξ ) = 2

12 = 1

6.

Now Pr(A) = 12 and Pr(B) = 2

6 = 1

3, so Pr(A) × Pr(B) = 1

2 × 1

3 = 1

6.

So Pr(A ∩ B) = Pr(A) × Pr(B); thus the two events are independent.

WorkEd ExamplE 29

Given that events A and B are independent, find the value of x if Pr(A) = 0.55, Pr(B) = 0.6 and Pr(A ∩ B) = x.

Think WriTE

1 Write the formula for independent events. Pr(A ∩ B) = Pr(A) × Pr(B)

2 Substitute the given information. x = 0.55 × 0.6

3 Simplify. x = 0.33

WorkEd ExamplE 30

Show that if Pr(A) = 0.5, Pr(B) = 0.8 and Pr(A ∪ B) = 0.9, then A and B are independent.

Think WriTE

1 Use the Addition Law for probabilities to find Pr(A ∩ B).

Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B) 0.9 = 0.5 + 0.8 − Pr(A ∩ B)Pr(A ∩ B) = 0.4

2 Calculate Pr(A) × Pr(B). Pr(A) × Pr(B) = 0.5 × 0.8 = 0.4Since Pr(A ∩ B) = Pr(A) × Pr(B), A and B are independent events.

WorkEd ExamplE 31

Two spinners each labelled with the numbers 1, 2, 3 are spun.A is the event ‘an odd number with the first spinner’.B is the event ‘an even number with the second spinner’.C is the event ‘an odd number from each spinner’.a Calculate Pr(A), Pr(B) and Pr(C ).b Decide which of the pairs of events AB, AC, BC is independent.

Think WriTE

a 1 List ξ, A, B and C. a ξ = {11, 12, 13, 21, 22, 23, 31, 32, 33}A = {11, 12, 13, 31, 32, 33}, B = {12, 22, 32},C = {11, 13, 31, 33}

2 Calculate Pr(A), Pr(B) and Pr(C ). Pr(A) = 69 = 2

3, Pr(B) = 3

9 = 1

3, Pr(C) = 4

9

Spinner 2Spinner 1

1 2

3

1 2

3


b Check to see if Pr(A ∩ B) = Pr(A) × Pr(B), Pr(A ∩ C) = Pr(A) × Pr(C ), Pr(B ∩ C) = Pr(B) × Pr(C ).

b A ∩ B = {12, 32}, Pr(A ∩ B) = 29

Pr(A) × Pr(B) = 23 × 1

3

= 29

Pr (A ∩ B) = Pr(A) × Pr(B), so A and B are independent.A ∩ C = {11, 13, 31, 33}, Pr(A ∩ C ) = 4

9

Pr(A) × Pr(C ) = 23 × 4

9

= 827

Pr(A ∩ C ) ≠ Pr(A) × Pr(C ), so A and C are not independent.B ∩ C = ϕ, Pr(B ∩ C ) = 0

Pr(B) × Pr(C ) = 13 × 4

9 = 4

27

Pr(B ∩ C ) ≠ Pr(B) × Pr(C ), so B and C are not independent.

When the probabilities of all possible outcomes are not equally likely, the probability of each outcome is placed on the corresponding branch of the tree diagram. When each branch is representing an outcome from independent events, you can follow the branches and multiply the probabilities together.

WorkEd ExamplE 32

A moneybox contains three $1 coins and two $2 coins. The moneybox is shaken; one coin falls out and is put back in the box. This is repeated twice more. If each coin has an equal probability of falling out:a represent this information on a tree diagramb calculate the probability of getting three $1 coinsc calculate the probability of getting at least two $2 coins.

Think WriTE/draW

a 1 There are three $1 coins and fi ve coins altogether. The probability of a $1 coin falling out is 3

5 = 0.6.

a

2 There are two $2 coins and fi ve coins altogether. The probability of a $2 coin falling out is 2

5 = 0.4.

3 Place the probability of each outcome on the corresponding branch of the tree diagram. $1

$2

$1

$2

$2

$1

$1 $1, $1, $1$2 $1, $1, $2

$1 $1, $2, $1$2 $1, $2, $2

$1 $2, $1, $1$2 $2, $1, $2

$1 $2, $2, $1$2 $2, $2, $2

0.6

0.4

0.6

0.40.6

0.4

0.6

0.40.6

0.4

0.6

0.4

0.6

0.4

b Multiply the probabilities obtained from the tree diagram corresponding to three $1 coins.

b Pr(three $1 coins) = 0.6 × 0.6 × 0.6 = 0.216

c 1 Outcomes corresponding to ‘at least two $2 coins’ are ($1, $2, $2), ($2, $1, $2), ($2, $2, $1) and ($2, $2, $2).

c


2 Calculate and add the probabilities. Pr(at least two $2 coins) = Pr($1, $2, $2 or $2, $1, $2 or $2, $2, $1 or $2, $2, $2)= (0.6 × 0.4 × 0.4) + (0.4 × 0.6 × 0.4) + (0.4 × 0.4 × 0.6) + (0.4 × 0.4 × 0.4)= 0.096 + 0.096 + 0.096 + 0.064= 0.352

WorkEd ExamplE 33

Christos estimates his chances of passing Maths, Science and English as 0.75, 0.6 and 0.5 respectively.a Represent this information on a tree diagram.b Assuming the events are independent, calculate the probability that: i he passes all three subjects ii he passes at least Maths and English iii he passes at least one subject.

Think WriTE/draW

a 1 Name the three events. a Let M, S, E be the events ‘passing Maths’, ‘passing Science’ and ‘passing English’ respectively.

2 Calculate Pr(M ′), Pr(S ′) and Pr(E ′). M ′ is the event of failing Maths.

Pr(M) = 0.75 Pr(M′) = 0.25 Pr(S) = 0.6 Pr(S′) = 0.4 Pr(E ) = 0.5 Pr(E′) = 0.5

3 Use the information to draw the tree diagram.

S

S'

S

S'

M'

M

E MSEE' MSE'

E MS'EE' MS'E'

E M'SEE' M'SE'

E M'S'EE' M'S'E'

Maths Science English

0.75

0.25

0.5

0.50.5

0.5

0.5

0.50.5

0.5

0.4

0.6

0.6

0.4

b i Multiply the probabilities corresponding to passes in all three subjects.

b i Pr(MSE) = Pr(M) × Pr(S) × Pr(E)= 0.75 × 0.6 × 0.5= 0.225

ii 1 We require that Christos pass both Maths and English and either pass or fail Science.

ii Pr(MSE or MS ′E)= Pr(MSE) + Pr(MS ′E)

2 These events are independent, so we may multiply the individual probabilities.

= Pr(M) × Pr(S) × Pr(E) + Pr(M) × Pr(S ′) × Pr(E)

= 0.75 × 0.6 × 0.5 + 0.75 × 0.4 × 0.5

3 Simplify. = 0.375

iii Passing at least one subject is the complement of failing all three subjects.

iii Pr(passes at least one subject) = 1 – Pr( M′S′E′)= 1 – 0.25 × 0.4 × 0.5= 1 – 0.05= 0.95



Exercise 11h independent events 1 WE29 Given that events A and B are independent, find the value of x if:

a Pr(A) = 0.4, Pr(B) = 0.5 and Pr(A ∩ B) = xb Pr(A) = 0.7, Pr(B) = x and Pr(A ∩ B) = 0.49c Pr(A) = x, Pr(B) = 0.8 and Pr(A ∩ B) = 0.32d Pr(A) = 0, Pr(B) = 0.5 and Pr(A ∩ B) = xe Pr(A) = 0.375, Pr(B) = x and Pr(A ∩ B) = 0.225.

2 WE30 Show that if Pr(A) = 0.6, Pr(B) = 0.25 and Pr(A ∪ B) = 0.7, then A and B are independent.

3 Two coins are tossed.a List the event space.b Show that the two events ‘Heads with the fi rst coin’ and ‘Tails with the second coin’ are

independent.

4 A coin is tossed twice. If A is the event ‘Heads with the first toss’ and B is the event ‘two Heads’, decide if the two events are independent.

5 A coin is tossed and a die is rolled.a What is the probability of getting Heads with the coin and a number greater than 2 with

the die?b Establish if the events ‘Tails with the coin’ and ‘getting an even number with the die’ are

independent.

6 mC Pr(A) = 0.4 and Pr(B) = 0.5. If A and B are independent events, the value of Pr(A ∪ B) is:a 0.5 B 0.7 C 0.4 d 0.9 E 0.8

7 WE31 A standard die coloured red and a standard die coloured blue are rolled. If A = ‘two odd numbers’, B = ‘a 1 or a 5 with the first die’ and C = ‘the sum of the two numbers is less than 4’:a calculate Pr(A), Pr(B) and Pr(C )b decide whether each of AB, AC and BC are independent.

8 mC Two coins are tossed and a die is rolled. The probability that there are less than two Heads and the number showing uppermost on the die is a 2 or a 5 is:a 3

4B 1

3C 2

3d 1

24 E 1

4

9 mC The probabilities of Anna, Bianca and Celia passing a Geography test are 0.75, 0.5 and 0.6 respectively. The probability that only two girls will pass the next Geography test is:a 0.65 B 0.275 C 0.45 d 0.14 E 0.15

10 mC A die is biased so that the probability of rolling a 1, 2, 3, 4, 5 or 6 is 0.25, 0.2, 0.1, 0.1, 0.15 or 0.2 respectively. If the die is rolled twice, the probability that the sum of the two numbers rolled is greater than 9 is:a 0.2255 B 0.4355 C 0.5650 d 0.1625 E 0.1255

11 WE32 A box contains 6 red marbles and 4 blue marbles. One marble is randomly drawn, its colour is noted and the marble is put back in the box. This procedure is done two more times. Represent the information as a tree diagram and calculate the probability of getting:a three red marblesb two red marbles and one blue marble in any orderc three red marbles or three blue marbles.

12 One card is randomly drawn from a standard deck of 52 cards, then the card is replaced and a second card randomly chosen. Determine the probability that:a both cards are acesb both cards are spadesc the two cards are different colours.

13 A die is biased so that the probability of obtaining the numbers 1, 2, 3, 4, 5 or 6 is 0.1, 0.3, 0.1, 0.2, 0.2 and 0.1 respectively. If the die is rolled twice, find the probability of rolling:a two 6sb an odd number followed by an even numberc two numbers that sum to 4d two numbers whose sum is greater than 10.


14 A Krisp-O cereal box contains an action card of a famous sports player selected from 10 cricket stars, 25 football identities and 15 tennis celebrities.a What is the probability that a randomly selected box of Krisp-O contains a card of a cricket

player?b What is the probability of fi nding cards of football players in each of two boxes randomly

selected from the supermarket shelf?c What is the probability that three randomly selected packs of Krisp-O will provide cards of

different sports?

15 WE33 Three types of seedling (daisy, rose and orchid) have probabilities of surviving any one week as 0.9, 0.85 and 0.8 respectively.a Represent this information on a tree diagram.b Assuming the seedlings’ chance of survival are independent, calculate the probability that: i after one week all three seedlings will have survived ii more than one seedling will survive to the end of the fi rst week.

16 Ibrahim estimates the probability of rain on Monday and Tuesday as 0.7 and 0.4 respectively. Assuming that the events ‘rain on Monday’ and ‘rain on Tuesday’ are independent:a represent this information on a tree diagramb calculate the probability of rain on both daysc calculate the probability of no rain on at least one of the two days.

17 A school junior swim team has five Year 7 students and seven Year 8 students. A merit certificate is to be awarded to one student from Year 7 and to one student from Year 8. a If each student has an equal chance of selection, state the probability of a particular student from

Year 7 and a particular student from Year 8 receiving the award.b If a special achievement certifi cate is also to be given to one of the students, what is the

probability that a particular student wins two awards, assuming that each student has the same chance of being selected?

18 To open a combination lock involves using the correct sequence of three digits selected from 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9, where a digit may be used more than once. If each digit is randomly selected:a calculate the probability of success after

one tryb calculate the probability of success after

one try given that the three digits are known to be odd

c calculate the probability of success after one try given that the fi rst and third digits are known to be the same.

19 A leather bag contains 4 black beads, 3 red beads and 3 white beads. Inside a plastic bag are 5 black beads, 2 red beads and 3 white beads. A nylon bag contains 6 black beads, 1 red bead and 3 white beads. One bead is randomly withdrawn from each bag. What is the probability of getting:a three black beads?b a red bead from the leather bag but not a red bead from the plastic bag?c at least two white beads?

20 One letter is randomly selected from each of the words HOORAY FOR MATHS. Find the probability of getting:a three vowelsb at least two consonantsc at least one vowel which is not the letter O.

21 At Greengate Secondary College, the probability of a VCE student proceeding to university studies is 50%, the probability of enrolling in TAFE courses is 20%, and there is a 30% probability of finding employment immediately after completion of the VCE. For a group of four randomly selected VCE students, what is the probability that:a all of them will undertake university studies?b all will seek employment or all will undertake TAFE courses?

diGiTal doCdoc-9809

SkillSHEET 11.2Sampling without

replacement


11i SimulationSimulation methods are used to model events when direct investigations may not be possible or practical because of factors such as insuffi cient time, possible danger or the expense involved. The aim of such methods is to obtain results comparable to the outcomes that would have been observed if the event had been examined directly. For example, the possible effects of air resistance on the structure and stability of a newly designed aeroplane may be investigated by performing wind-tunnel experiments using a scale model of the aircraft. On a larger scale, the economic implications arising from interest rate changes may be investigated using mathematical models that will take into account relevant variables such as unemployment and the cost of living. Altering the value and type of variable will provide a range of predicted outcomes.

Many basic simulation techniques involve the generation of random numbers. Methods used include coins, dice, cards, spinners, random number tables, calculators and computers. A CAS calculator can be especially useful.

WorkEd ExamplE 34

Pina estimates her chance of passing an English test as 23, and must take 5 tests during the year.

a Explain how a CAS calculator may be used to simulate the 5 tests.b Carry out a simulation to obtain an estimate of Pina’s performance on the 5 tests.

Think WriTE/diSplaY

a 1 Defi ne the event as shown. a Let P be the event ‘passing the test’.

2 State the relevant probabilities. Pr(P) = 23 and Pr(P′) = 1

3

3 Decide which numbers (to be generated on a CAS calculator) will represent the event space and which will represent favourable outcomes.

Let the numbers 1, 2, 3 represent the event space, 1 and 2 represent passing the test and 3 represent failing the test.

b 1 Use the random number generator feature of the CAS calculator to generate 5 random numbers from the set {1, 2, 3}.

b randInt(1, 3, 5)

2 Note that the results will vary every time. {3, 3, 1, 2, 2} is a possible set of 5 random numbers.

3 Three of the 5 outcomes (1, 2 and 2) represent passing the test.

The simulation predicts Pina will pass 3 of the 5 tests.The results can be summarised as follows.

TrialRandom number

Test outcome

1 3 Fail

2 3 Fail

3 1 Pass

4 2 Pass

5 2 Pass

WorkEd ExamplE 35

The probability of a Jonathan apple tree producing fruit in any one season is 56 and the

probability of a Granny Smith apple tree bearing fruit in a season is 34.

a Assuming the two events are independent, calculate the probability that during a particular season:

i both trees will produce fruit ii both trees will bear no fruitiii only one of the trees will bear fruit.

b Devise a suitable simulation model consisting of 10 trials for each tree to obtain estimates for the probabilities obtained in a.

diGiTal doCdoc-9810ExtensionSampling without replacement


Think WriTE

a 1 Defi ne the events as shown. a Let J = Jonathan bears fruit G = Granny Smith bears fruit

2 State the given probabilities. Pr(J ) = 56, Pr(G) = 3

4

i Find Pr(Jonathan and Granny Smith bear fruit). Recall ∩ means ‘and’. Since the events are independent, Pr(J ∩ G) = Pr(J) × Pr(G).

i Pr(J ∩ G) = 56 × 3

4

= 58

ii Find Pr(both trees bear no fruit).Note: Pr(J ′ ) = 1 − Pr(J ) = 1 − 5

6 = 1

6 and

Pr(G′ ) = 1 − Pr(G) = 1 − 34 = 1

4

ii Pr(J′ ∩ G′) = 16 × 1

4

= 124

iii Consider all situations in which only one of the trees bears fruit and add the probabilities (recall ‘or’ means + ).

iii Pr(one tree bears fruit)= Pr(J ∩ G′) + Pr(J′ ∩ G)

= 56 × 1

4 + 1

6 × 3

4

= 13

b 1 Decide on a method of simulation and decide which numbers will represent the event space and which will represent favourable outcomes. There are 6 possible outcomes when a die is rolled: 1, 2, 3, 4, 5, 6.There are 4 possible outcomes when two coins are tossed: HH, TT, HT, TH.

b Let rolling a 1, 2, 3, 4 or 5 on a die represent the Jonathan bearing fruit.Let the outcomes HH, TT and HT when two coins are tossed represent the Granny Smith bearing fruit.

2 Roll a die 10 times and record the results. Possible results are shown in the table.Alternatively, use a CAS calculator to generate 10 random numbers between 1 and 6 by entering (1, 6, 10) in randInt.

✓ = Fruit × = No fruitTrial Die Coin Jonathan Granny Smith

12345678910

3412653611

THHTTTHHTHHTTHHHHTHH

✓✓✓✓✕✓✓✕✓✓

✕✓✓✓✕✓✕✓✓✓

3 Toss two coins 10 times and record the results. Possible results are shown in the table.Alternatively, use a graphics calculator to generate 10 random numbers between 1 and 4 by entering (1, 4, 10) in randInt.

i Trials 2, 3, 4, 6, 9 and 10 (6 trials) correspond to both trees bearing fruit.

i Estimated probability that both trees bear fruit = 6

10

ii Only trial 5 corresponds to both trees bearing no fruit.

ii Estimated probability that both trees bear no fruit = 1

10

iii Trials 1, 7 and 8 correspond to only one tree bearing fruit.

iii Estimated probability that only one tree bears fruit = 3

10

Compare the calculated and simulated probabilities in worked example 35. Repeat the simulation to obtain a new set of results. How do they compare?


Exercise 11i Simulation 1 WE34 A student estimates the probability of stopping at a particular set of traffic lights when being

driven to school in the morning is 15.

a Explain how a graphics calculator can be used to simulate 5 trips to school.b Carry out 5 simulations to obtain an estimate for the probability of stopping at the intersection.

2 The table below shows the number of bullseyes scored by 40 dart players after 5 throws each.

1 4 0 3 4 2 1 5 4 23 0 4 5 2 1 4 3 2 10 2 1 4 3 5 3 2 4 40 2 1 0 3 5 4 2 3 1

a Explain how a die or a CAS calculator can be used to obtain the range of numbers given in the table.

b What proportion of players scored at least 3 bullseyes?c Using a die (or by some other means), conduct 20 trials and obtain a second value for b.d Analysis of the results of a particular tournament (at which each player threw 3 darts) gives

the probability of scoring more than 2 bullseyes to be twice the probability of scoring less than 3 bullseyes.

i Explain how a die or a CAS calculator can be used to simulate this situation. ii Perform 40 trials and compare your results to the given probabilities.

3 A student generated 30 three-digit random numbers using a calculator. The results are given in the table below.

200 123 399 165 100 355 778 400 150 100

387 001 793 215 030 288 345 009 970 993

546 720 549 139 248 405 369 217 935 782

a Explain how you can obtain 30 random 2-digit numbers from the table to simulate the ages of 30 people aged from 10 years to 100 years.

b Using the method you suggested for a, obtain an estimate of the proportion of the group that is younger than 50 years of age.

4 WE35 A car may be said to be safe if both of its airbags will operate properly in the event of a collision. Suppose that the probability of the driver’s airbag failing is 1

4 and the probability of the front

passenger’s airbag not working is 13. Use one die and two coins to simulate the situation of the airbags

working/not working.a Assuming the operation of the airbags to be

independent, fi nd the probability that during a collision:

i both airbags operate properly ii neither airbag operates properly iii only one airbag operates properly.b Using a die and two coins, devise a simulation

model consisting of 20 trials to obtain an estimate for the probabilities obtained in a.

5 A die and a coin are each tossed 20 times with the following results.

Die 1 3 2 4 6 5 4 3 2 3 1 6 4 6 2 1 4 2 2 1

Coin H H T H T T T H H T H T H T H T H T H H

H is Heads, T is Tails.

a Explain how the events of tossing a coin and rolling a die can be used to simulate the situation of forming a particular dancing couple consisting of a man randomly chosen from a group of 2 men and a woman randomly selected from a group of 3 women.

b Perform 20 simulations to obtain an estimate of the probability of a particular man and a particular woman dancing together.

6 Each packet of Krisp potato chips contains 1 of 5 different fridge magnets. Use 20 trials for a simulation analysis to determine how many packets of Krisps need to be purchased in order to obtain all 5 magnets.

diGiTal doCdoc-9811random numbers


Summaryintroduction to experimental probability

Sets and Venn diagrams:• { } A set is a collection of objects (or elements).• x ∈ A denotes that element x belongs to set A.• x ∉ A denotes that element x does not belong to set A.• A fi nite set can be listed; an infi nite set cannot be listed.• A null set (denoted ϕ) contains no elements.• A unit set contains one element.• n(A) The cardinal number of a set is the number of elements it contains.• ξ The universal set is a set containing all elements being considered.• A′ The complement of set A is all the elements of the universal set not contained in set A.• A = B Two sets are equal if they contain the same elements.• A ↔ B Two sets are equivalent if they have the same cardinal number.• A ∩ B The intersection of two sets is the set of elements common to both sets.• A ∩ B = ϕ Two sets are disjoint sets if they have no elements in common.• A ∪ B The union of two sets A, B is the set that contains all elements belonging to A or B or to

both A and B.• n(A ∪ B) = n(A) + n(B) − n(A ∩ B)• A ⊂ B Set A is a subset of set B if all the elements of set A are contained in set B.• A ⊃ B Set A contains set B if set A contains all the elements of set B.• Properties of sets can be represented as Venn diagrams.

Using sets to solve practical problems:• Use overlapping regions (for example, circles) to represent sets within a universal set (for example,

a rectangle).• Label each set.• Fill in any given information.• Calculate required missing quantities.

Estimated probability and expected number of outcomes:

• Experimental probability = number of favourable outcomes observed


• Expected number of favourable outcomes = experimental probability (long-run proportion) × number of trials

Calculating probabilities • Pr(favourable outcome) =

number of favourable outcomes

total number of outcomes

• For event E, Pr(E ) = n(E )

n(ξ ).

• Pr(E ) + Pr(E ′) = 1• Pr(E ′) = 1 − Pr(E )• If event E is impossible, Pr(E ) = 0.• If event E is certain to occur, Pr(E ) = 1.• 0 ≤ Pr(E ) ≤ 1• Pr(ξ) = 1 and Pr(ϕ) = 0

Tree diagrams and lattice diagrams

• Outcomes can be illustrated by a tree diagram. The order of events determines the structure of the ‘tree’.• A lattice diagram is a grid used to show the possible outcomes when two events occur. It is

particularly useful when dealing with outcomes from rolling a die.

The addition law of probabilities

• For events A, B, Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B).• ∪ means ‘or’ and ∩ means ‘and’.• Two events are mutually exclusive if they cannot occur at the same time.• If A ∩ B = ϕ, then A, B are mutually exclusive and Pr(A ∩ B) = 0.


karnaugh maps and probability tables

• Karnaugh maps and probability tables summarise all combinations of two events (for example, A and B) and their complements (A′ and B′ ).

• Use known values, subtotals and the fact that the sum of the probabilities in the outer row and column is equal to 1 to fi ll in any missing values.

Conditional probability • For two events A and B, the conditional probability of event A given event B occurs is

Pr(A | B) = Pr(A ∩ B)

n(ξ ), where Pr(B) ≠ 0 and B is the reduced event space.

• Venn diagrams, tree diagrams and Karnaugh maps are useful aids in conditional probability problems.

• The Law of Total Probability: Pr(A) = Pr(A | B) × Pr(B) + Pr(A | B′) × Pr(B′)

Transition matrices and markov chains

• In general, Sn + 1 = T × Sn may be used to determine any state from the previous one.• In general, Sn = T n × S0.

independent events • If Pr(A ∩ B) = Pr(A) × Pr(B), then events A and B are independent.• Pr(A ∩ B) means the probability of events A and B.

Simulation • Simulation techniques are used to model events.• Outcomes of the event space are randomly obtained but based on certain probabilities.• Coins, dice, random number tables and calculators can provide random numbers that can be

associated with outcomes of the event space.• Performing more trials or repeating the simulation will provide better estimates of the probability.

AA'

B'B

1


Chapter review 1 A random sampling of 80 ceramic tiles produced at a ceramics factory reveals 8 scratched, 3 chipped

and 4 broken tiles. Estimate the probability that a tile produced at the factory will be:a scratchedb chipped or brokenc damaged in some way.

2 A card is randomly selected from a deck, its suit is noted, and then the card is placed back in the deck. The experiment is repeated to obtain a second card.a List the possible outcomes for selecting 2 cards in this way.b Are the outcomes all equally likely? Explain.

3 A game show host spins the wheel shown at right. What is the probability that the wheel ends on:a the jackpot ($1000 prize)?b a prize greater than $50?

4 A letter is chosen at random from each of the words GO BLUES. Represent all possible outcomes on a tree diagram and find the probability that:a G and B are chosenb S is chosenc G or S is chosen.

5 A standard die is thrown and the spinner shown at right is spun.a Show all possible outcomes on a lattice diagram.b Find the probability of getting a number greater than 4 on the die and

an odd number on the spinner.

6 A set of 20 uniformly sized cards numbered 1 to 20 is shuffled. What is the probability of drawing a number less than 8 or an even number from this set?

7 A class of 30 students was asked if there was a pet dog at home and if the students were responsible for pooper scooping before the backyard lawn was mown. Fourteen students had a dog but only 6 did the pooper scooping.a Draw a Karnaugh map showing this information.b Complete a probability table.c State the probability that a randomly selected student has a dog but avoids pooper scooping.

8 If Pr(A) = 0.3 and Pr(B | A) = 0.4, find Pr(A ∩ B).

9 Two identical, equally accessible cookie jars sit on a kitchen bench. Jar 1 contains 6 chocolate and 9 plain biscuits, and jar 2 contains 12 chocolate and 8 plain biscuits. One biscuit is selected randomly from one of the jars. If a chocolate biscuit is selected, what is the probability that it came from jar 1?

10 Of 50 people surveyed, 35 played tennis and 26 played netball. Everyone surveyed played at least one of these sports.a How many people played both netball and tennis?b If one person is selected at random, what is the probability that: i he/she plays tennis only? ii he/she plays netball? iii he/she plays tennis, given that he/she also plays netball?

mUlTiplEChoiCE

1 Twelve nuts are taken from a jar containing macadamias and cashews. If 3 macadamias are obtained, the estimated probability of obtaining a cashew is:

a 112

B 14

C 13

d 34

E 31

2 From a normal pack of 52 playing cards, one card is randomly drawn and replaced. If this is done 208 times, the number of red or picture cards (J, Q or K) expected to turn up is:a 150 B 130 C 120 d 160 E 128

3 A cubic die with faces numbered 2, 3, 4, 5, 6 and 6 is rolled. The probability of rolling an even number is:

a 13

B 23

C 16

d 56

E 12

ShorT anSWEr

$1000Jackpot

$60

Lose

$100$100

Boody

Prize

$40$500$30

$50

$200

Lose

it al

l

$10

23

1


4 The probability of rolling an odd number or a multiple of 2 using the die in question 3 is:

a 1 B 13

C 14

d 34

E 23

5 The tree diagram that describes the outcomes when three coins are tossed is:

a

H

TH

TH

TTH

TH

B H

H

HT

T

T

C

H

H

T

H

T

T HTHT

T

H

TH

d H

H H

TH

TT

T

H

T

E HT

HT

TH

6 Bag A contains 3 red and 4 blue marbles, and Bag B contains 3 yellow and 2 green marbles, as shown at right. A marble is drawn from Bag A, then one is taken from Bag B. Which diagram below best illustrates this situation?

a Yellow RY

Green RG

Yellow BY

Green BGBlue

RedB Red YR

Blue YB

Red GR

Blue GBGreen

Yellow

C Blue RB

Green RG

Blue YB

Green YGYellow

Redd

Green

Red

Blue

Yellow

E

Green

Red

Blue

Yellow

Green RG

Red RRBlue RBYellow RY

Green BG

Red BRBlue BBYellow BY

Green YG

Red YRBlue YBYellow YY

Green GG

Red GRBlue GBYellow GY

Bag A Bag B


7 If Pr(A) = 0.6, Pr(B) = 0.7 and Pr(A ∪ B) = 0.8, then Pr(A ∩ B) is:a 0.1B 0.5C 0.9d 0.2E 0.6

8 Consider the Karnaugh map at right, where C is ‘people who like comedy movies’ and A is ‘people who like action movies’.The number of people who like only action movies is:a 26 B 30 C 18d 20 E 23

9 Which of the following alternatives gives the correct values of a and b in the probability table at right?a a = 0.2, b = 0.3B a = 0.3, b = 0.2C a = 0.3, b = 0.6d a = 0.6, b = 0.3E a = 0.7, b = 0.4

10 If Pr(B | A) = 0.45 and Pr(A ∩ B) = 0.35, then Pr(A) is:

a 79

B 38

C 49

d 58

E 89

11 A fair coin is tossed twice. If A = Tails on first toss, B = Heads on second toss and C = both tosses are Heads, which of the following is true?a Pr(A) = 1

4B Pr(B) = Pr(C)C A and B are independent.d A and C are independent.E B and C are independent.

12 An archer has a probability of approximately 38 of hitting the bullseye from a particular distance. The

spinner at right is used to simulate 10 rounds of 4 shots at such a target, and the results are as follows, where B = blue (bullseye) and R = red (non-bullseye).B, R, R, R B, R, B, R R, B, R, RR, B, R, R R, B, B, RR, R, B, B R, B, R, R R, R, R, RR, B, B, R B, R, B, BBased on this simulation, the probability of getting 2 bullseyes in a round of 4 shots is:

a 14

B 38

C 25

d 58

E 34

ExTEndEd rESponSE 1 For a transition matrix T = 0.22 0.33

0.78 0.67

and an initial state matrix S0 = 0.5

0.5

, calculate, accurate

to 3 decimal places:a S1

b S4.

2 The manager of a snow resort has noticed that, if it snows on a given day, there is a 70% chance that it will snow the following day. If it does not snow, there is only a 30% chance that it will snow the following day. John arrived on a Saturday when the weather was sunny and clear.a What is the probability that he will have fresh snow the following Tuesday?b What is the probability that he will have fi ne weather for the drive home on the following

Saturday? (Give answers to 3 decimal places.)

3 A factory has a machine in poor working condition that often produces faulty components. If it produces a faulty component, there is a probability of 0.25 that it will follow this up with another faulty component. However, each time it produces a good component, there is a probability of only 0.05 that

A A′C 19

C′ 25

26 50

B B′A 0.4 0.7

A′ a b

0.4


it will next produce a faulty component. If there is a 20% chance that the first component of the day is faulty, set up the initial state matrix and find the probability that:a the second component is faulty b the fi fth component is faulty.

4 One student is selected at random from each of Years 7, 8 and 9. If there are 148 girls and 114 boys in Year 7, 126 girls and 97 boys in Year 8, and 115 girls and 122 boys in Year 9, find the probability that all students chosen are boys. Give your answer to 3 decimal places.

5 The probability that the newspaper is delivered to Geoff’s house before 8.00 am is 57, and the probability

that he arrives at work to fi nd a free parking space in his company’s car park is 910

. Assuming these events are independent:a use a CAS calculator to conduct a simulation of 28 days’ durationb fi nd the probability, based on the simulation, that on one day Geoff misses out on a company car

park space and his paper arrives latec calculate the theoretical probability of the combination of events described in part b.

6 Consider 3 fair coins being tossed.If A = Heads with the fi rst coin, B = Tails with the second coin and C = Tails with the third coin:a list the event space (for example, use HTH for

‘Head then Tail then Head’)b fi nd: i Pr(A) ii Pr(B) iii Pr(C) iv Pr(A ∩ B) v Pr(A ∩ C) vi Pr(B ∩ C) vii Pr(A ∩ B ∩ C) viii Pr(A) × Pr(B) × Pr(C)c propose how you might defi ne independence for 3 events.


7 A large number of asthma sufferers were asked to volunteer for the testing of a new drug. Only some of the volunteers were given the drug, but all of the volunteers were observed to see if they developed asthma on a smoggy day. The results are shown in the table below.

Given drug Not given drug

Developed asthma 148 59

Did not develop asthma

566 184

a How many people were selected to take part in the study?b If a person was randomly selected from the volunteers, what is the probability that they were

given the drug?c What is the probability that a randomly selected person developed asthma on this day?d Given that a volunteer was given the drug, what is the probability that they developed asthma?e From this information, what can you conclude about the effectiveness of the drug in preventing

asthma?It was decided that conclusive observations about the effectiveness of the drug could not be made after one day, so the same volunteers continued with the study for three months. (Assume that the number of people given the drug is unchanged.) The results were:

395 people were given the drug and did not have an asthma attack143 people were given the drug but had exactly one episode of asthma per month97 people were not given the drug and developed asthma more than once a month84 people were not given the drug and had exactly one episode of asthma each month.

f Represent this information using a Venn diagram.g How many people were given the drug and had more than one episode of asthma per month?h How many people in the study were not given the drug and did not have an episode of asthma?i Given that a volunteer had been given the drug, what is the probability that they have more than

one episode of asthma per month?j Given that a volunteer had more than one episode of asthma per month, what is the probability

they had taken the drug?

diGiTal doCdoc-9812

Test YourselfChapter 11


ICT activitiesChapter openerdiGiTal doC

• 10 Quick Questions doc-9801: Warm-up with ten quick questions on introductory probability (page 475)

11a introduction to experimental probabilitydiGiTal doCS

• doc-9802: Investigate long-run proportion using a spreadsheet (page 475)

• doc-9803: Simulate the rolling of a die multiple times using a spreadsheet (page 475)

• doc-9804: Simulate the rolling of two dice multiple times using a spreadsheet (page 475)

• WorkSHEET 11.1 doc-9805: Calculate probabilities for everyday events and random selection (page 478)

11B Calculating probabilitiesdiGiTal doC

• Extension doc-9810: Investigate sets and Venn diagrams (page 478)

11C Tree diagrams and lattice diagrams

TUTorial• WE 9 eles-1448: Assuming replacement, use a tree diagram to

calculate the probabilities of drawing two specific cards (page 483)

diGiTal doC• doc-9806: Investigate Stirling’s formula using a spreadsheet

(page 485)

11d The addition law of probabilities

TUTorial• WE 15 eles-1449: Use the addition law of probabilities to

calculate probabilities and determine whether two events are mutually exclusive (page 488)

diGiTal doC• WorkSHEET 11.2 doc-9807: Use Venn diagrams to help calculate

probabilities (page 492)

11E karnaugh maps and probability tables

TUTorial• WE 18 eles-1450: Complete a probability table and use it to

represent the information in a Venn diagram (page 493)

11F Conditional probabilityTUTorial

• WE 24 eles-1451: Use a tree diagram to calculate a conditional probability (page 499)

diGiTal doC• SkillSHEET 11.1 doc-9808: Practise conditional probability

(page 500)

11G Transition matrices and markov chainsinTEraCTiViTY

• Transition matrices int-0270: Consolidate your understanding of transition matrices (page 501)

TUTorial• WE 25 eles-1452: Represent conditional probabilities of two

events in a table (page 503)

11h independent eventsTUTorial

• WE 33 eles-1453: Using a tree diagram, calculate probabilities assuming three events are independent (page 510)

diGiTal doC• SkillSHEET 11.2 doc-9809: Practise sampling without replacement

(page 512)

11i SimulationdiGiTal doCS

• Extension doc-9810: Investigate sampling without replacement (page 513)

• doc-9811: Investigate random numbers using a spreadsheet (page 515)

Chapter reviewdiGiTal doC

• Test Yourself doc-9812: take the end-of-chapter test to test your progress (page 522)

To access eBookPLUS activities, log on to www.jacplus.com.au


Answers CHAPTER 11

inTrodUCTorY proBaBiliTY Exercise 11a introduction to experimental probability 1 0.6 2 0.7 3 a 0.66 b 0.34 4 48 5 250 6 200 7 A 8 C 9 18 squares, 24 circles 10 a 120 b 140 11 E 12 249 (Hot-Shot), 401 (Zap Inc) 13 10 14 a i 72 ii 24 iii 12 b 4 losses or draws 15 A 16 a 0.7597 b 0.2403 c i 1200 ii 380

Exercise 11B Calculating probabilities 1 a ξ = {Red, Blue, Yellow, Green} b n(ξ) = 4 2 ξ = {A, B, C, D, E, F, G, . . . , X, Y, Z} 3 n(ξ) = 52 4 a ξ = {spades H, spades T, clubs H,

clubs T, hearts H, hearts T, diamonds H, diamonds T}

b S = {spades H, spades T} 5 ξ = {HH, HT, TH, TT} 6 a ξ = {boy Time Out, boy Mars Bar, boy

Violet Crumble, girl Time Out, girl Mars Bar, girl Violet Crumble}

b M = {boy Mars Bar, girl Mars Bar}

7 a 1

7 b

6

7

8 a 1 b 1

154 c

73

154

9 a 1

13 b

1

4 c

3

13

d 10

13 e 1

10 1

40000

11 a 1

2 b

1

6 c

1

2 d 1

12 a 1

2 b

4

5 c

1

4 d

1

2

13 a 5

11 b

4

11

14 9

21

3

7=

15 a 3

19 b 7

19 c 12

19 d

14

19

16 a 8

17 b

1

85 c 0

Exercise 11C Tree diagrams and lattice diagrams 1 a B

RGBRGBRG

BBBRBGRBRRRGGBGRGG

B

R

G

n( ) = 9ξ

2 a Red HR

White HW

Red TR

White TWTail

Head

n( ) = 4ξ

b 1

4

3 a A CA

T CT

C TC

A TAT

C

AC AC

T AT

b 1

6

4 a H HH

T HT

H TH

T TTT

H b

1

2

5 a 1

4 b

1

4 c

1

4

6 B 7

4

6

2

8

10

Red 2RBlue 2BRed 4RBlue 4BRed 6RBlue 6BRed 8RBlue 8BRed 10RBlue 10B

n( ) = 10ξ

3

10

8 456456

2

3

2 42 52 63 43 53 6

n( ) = 6ξ

9 1

12

n(ξ) = 12J1 is the fi rst jacket,T1 is the fi rst tie.

J1

T1T2T3T4

J2

T1T2T3T4

J3

T1T2T3T4

10

T4

T1

T2

T3

Outcomes

B

R

G

B

R

G

B

R

G

B

R

G

LFLFLFLFLFLFLFLFLFLFLFLF

n( ) = 24T1 is the �rst small triangle.R is red, G is green, B is blue.L is low sheen.F is full gloss.

ξ

1

6

11 a Head

Tail

Head

TailTail

HeadHead HHHTail HHTHead HTHTail HTTHead THHTail THTHead TTHTail TTT

n( ) = 8ξ

b 1

8 c

1

2

12

B

V

L

S

C

C

C

C

S VCSL VCLB VCBV LCVS LCSB LCBV SCVL SCLB SCBV BCVS BCSL BCL

Outcomes n(ξ) = 12

13 Alan

Bjorn

Carl

Head AHTail ATHead BH

Tail BTHead CHTail CT

n( ) = 6ξ

1

3

14 a

S

M

T

H

S

S

S

S

F MSFMSNTSFTSNHSFHSNSSFSSN

NFNFNFN

b 1

4

15 a 1

4 b

1

4 c

1

4

16 3

8 17 D

18 a

H

T

H

T H

T

H

T

HT

H

T

T

H

H HHHHT HHHTH HHTHT HHTTH HTHHT HTHTH HTTHT HTTTH THHHT THHTH THTHT THTTH TTHHT TTHTH TTTHT TTTT

Outcomes

b 1

16 c

3

8 d

1

2

19 a

Die 1

Die

2

123456

1 2 3 4 5 6

b 1

4 c 1

9

20 a

Die

Coi

n

HT

1 2 3 4 5 6

b 1

12 c

1

4

21 a 1

36 b

1

18 c

5

6

d 1

9 e

1

4

b i 1

9

ii 1

9

iii 2

9

a 1

3

b 1

6

c 1

2


22 16

23 M1

M2

M1

M2

M1

M2E3

E1

E2

D1D2D1D2D1D2D1D2D1D2D1D2

E1 is the �rst entrée. M1 is the �rst main meal.D1 is the �rst dessert.n( ) = 12ξ

1

6

Exercise 11d The addition law of probabilities 1 0.7 2 0.68 3 0.89 4 0.3 5 0.45 6 0.41 7 0.78, 0.39 8 A and B are mutually exclusive. 9 A ⊂ B 10 a i

1

4 ii

3

13 iii

3

52 iv

11

26

b No

11 a i 3

5 ii

2

5 iii 0 iv 1

b Yes 12 a

1

10 b

1

2 c

3

20 d

1

2

13 a 1

2 b

3

10 c

13

20

14 2

3 15

2

5

16 a 1

13 b

1

4

c 2

13 d

27

52

17 a 27

80 b

3

8 c 12

65

18 a 1

14 b

1

7

c 2

7 d

9

14

19 a 19

24 b

5

8

c 13

24 d

11

24

20 a 7

10 b

2

5

c 1 d 4

5

21 a 3

4 b

23

40

c 29

40 d 0

22 The probability of winning Alotto, Blotto

or Clotto is 33

100,

30

100 and

32

100 respectively, so

Alotto is easier to win. 23 0.75

24 3

10

25 a i 1

10 ii

2

5

iii 1

2 iv

9

10

b The areas are stated to be equal to ensure that each area has an equal chance of being hit, that is, to ensure equally likely outcomes.

Exercise 11E karnaugh maps and probability tables 1 a

B B′A 8 17 25

A′ 7 6 13

15 23 38

ξA B

17 8 7

6

b B B′A 69 33 102

A′ 45 27 72

114 60 174

ξA B

33 69 45

27

c B B′A 0.27 0.3 0.57

A′ 0.4 0.03 0.43

0.67 0.33 1

ξ

A B

0.3 0.27 0.4

0.03

d B B′A 0.61 0.03 0.64

A′ 0.14 0.22 0.36

0.75 0.25 1

ξA B

0.03 0.61 0.14

0.22

2 C 3

B B′A 87 63 150

A′ 13 55 68

100 118 218

4 B B′A 35 29 64

A′ 56 44 100

91 73 164

5 a B B′A 0.45 0.14 0.59

A′ 0.3 0.11 0.41

0.75 0.25 1

b B B′A 0.61 0.27 0.88

A′ 0.12 0 0.12

0.73 0.27 1

c B B′A 6 10 16

A′ 4 0 4

10 10 20

d B B′A 7 15 22

A′ 18 5 23

25 20 45

6 a B B′A 0.27 0.37 0.64

A′ 0.09 0.27 0.36

0.36 0.64 1

b B B′A 0.15 0.35 0.5

A′ 0.25 0.25 0.5

0.4 0.6 1

7 E 8 a ξ

F S

24 6 20

10

b S S′F 6 24 30

F′ 20 10 30

26 34 60

9 S S′F 0.1 0.4 0.5

F′ 0.33 0.17 0.5

0.43 0.57 1

F is football, S is soccer.


10 A 11 a B B′

A 0.3 0.3 0.6

A′ 0.2 0.2 0.4

0.5 0.5 1

b B B′A 0.1 0.1 0.2

A′ 0.7 0.1 0.8

0.8 0.2 1

c B B′A 0.5 0.3 0.8

A′ 0.1 0.1 0.2

0.6 0.4 1

d B B′

A 0 1

4

1

4

A′ 1

4

1

2

3

4

1

4

3

41

12 S S′C 50 110 160

C′ 95 25 120

145 135 280 C is chocolate, S is strawberry.

13 a G G′A 0.32 0.22 0.54

A′ 0.44 0.02 0.46

0.76 0.24 1

A is people less than 20 years of age, G is people who wear glasses.

b i 0.02 ii 0.54 14 a B B′

A 0.08 0.32 0.4

A′ 0.25 0.35 0.6

0.33 0.67 1

b i 0.35 ii 0.25 iii 0.57

15 a B B′D 0.18 0.42 0.6

D′ 0.28 0.12 0.4

0.46 0.54 1 D is Daily Times, B is Bugle.

b i 0.4 ii 0.28 iii 0.54 c 132 16 a

B B′

R 10% 15% 25%

R′ 10% 65% 75%

20% 80% 100%

b 0.15 c 0.75

Exercise 11F Conditional probability 1 a 0.8 b 0.5 2 a 0.6 b 0.69 3

4

5 4

5

12 5 0.144

6 0.168 7 2

3 8

5

6

9 a 0.3 b 3

7

10 D 11 A

12 B AB A

A

A B

AB

Pr( | )Pr( )

Pr( )

Pr( ) Pr( )

Pr( )Pr( )

= ∩ = ×

=

13 a 0.2 b 20

27

14 a 0.3 b 2

3 c ξ

A B

0.15 0.3 0.22

0.33

15 a 3

10 b

19

30 16 A 17 a 15 b ξ Mountain Scuba

15 15 30

c 1

2 d

1

2

18 11

14 19

2

7

20 a 2

3 b

2

75 c

2

25

21 a 0.40 b 0.32 c 0.13 d 0.21 e 0.63

22 6

13

Exercise 11G Transition matrices and markov chains

1 0.75 0.50.25 0.5

2 0.31

3 a 6733

b

63.336.7

c 63.6736.33

4 a 380620

b

443557

c 435565

5 0.411 0.3681250.589 0.631875

0.4110.589

6 0.772 694 0.757 6880.227 306 0.242 313

0.772 6940.227 306

7 a 0.70 0.600.30 0.40

b

200300

c Vegetable 333, tomato 167 d Same answer as c 8 a 0.80 b 0.58 9 a i 0.7500 ii 0.6880 iii 0.6875 iv 0.6875 b i 0.5500 ii 0.6864 iii 0.6875 iv 0.6875 c 0.6875

10 a 0.40 0.750.60 0.25

b 0.5365

c 0.4635 d 0.4444

Exercise 11h independent events 1 a 0.2 b 0.7 c 0.4 d 0 e 0.6 2 Using the Addition Law, Pr(A ∩ B) = 0.15

and Pr(A) × Pr(B) = 0.15, so A and B are independent.

3 a {H1H2, H1T2, T1H2, T2T2} b Pr(H1 ∩ T2) = 1

4 and Pr(H1) × Pr(T2)

1

4=

4 Not independent 5 a 1

3 b Independent

6 B 7 a

1

4,

1

3,

1

12

b A and B are not independent. A and C are not independent. B and C are not independent.

8 E 9 C 10 D 11

First drawSecond draw Third draw

Red

Blue

Red

BlueBlue

Red

RedBlueRedBlueRedBlueRedBlue

0.6

0.4

0.6

0.4

0.6

0.40.6

0.40.6

0.40.6

0.4

0.6

0.4

a 0.216 b 0.432 c 0.28

12 a 1

169 b

1

16 c 1

2

13 a 0.01 b 0.24 c 0.11 d 0.05

14 a 1

5 b

1

4 c

9

50

15 aDaisy Rose Orchid

Survive

Notsurvive

Survive

Notsurvive

Notsurvive

Survive0.9

0.1

0.85

0.15

0.85

0.15

0.8

0.2

SurviveNot survive

0.8

0.2

SurviveNot survive

0.8

0.2

SurviveNot survive

0.8

0.2

SurviveNot survive

b i 0.612 ii 0.941 16 a Monday Tuesday

No rain

Norain

RainRain

No rainRain

0.7

0.3

0.4

0.6

0.4

0.6

b 0.28 c 0.72


17 a 1

35 b 1

35

18 a 1

1000 b

1

125

c 1

100

19 a 3

25 b

6

25

c 27

125

20 a 1

30 b

11

15

c 1

3

21 a 0.0625 b 0.0097

Exercise 11i SimulationAnswers will vary.

ChapTEr rEViEWShorT anSWEr

1 a 1

10 b 7

80 c 3

16

2 a Let H = hearts, D = diamonds, S = spades and C = clubs. Possible outcomes are:HH DH SH CHHD DD SD CDHS DS SS CSHC DC SC CC

b All outcomes are equally likely as there is an equal number of cards in each suit during each selection.

3 a 1

12 b

5

12

4 BLUESBLUES

GBGLGUGEGSOBOLOUOEOS

G

O

a 1

10 b

1

5 c

3

5

5 a

Die

Spin

ner

123

1 2 3 4 5 6

b 2

9

6 7

10

7 a P P′D 6 8 14

D′ 0 16 16

6 24 30

b P P′

D 3

15

4

15

7

15

D′ 08

15

8

15

3

15

12

151

c 4

15

8 0.12

9 2

5

10 a 11

b i 12

25 ii

13

25 iii 11

26

mUlTiplE ChoiCE

1 D 2 E 3 B 4 A 5 C 6 A 7 B 8 D 9 A 10 A 11 C 12 C

ExTEndEd rESponSE

1 a 0.2750.725

b

0.2970.703

2 a 0.468 b 0.501

3 S0 0.20.8

a 0.09 b 0.063

4 0.097 5 a, b Answers will vary.

c 1

35

6 a {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

b i 1

2 ii

1

2

iii 1

2 iv

1

4

v 1

4 vi

1

4

vii 1

8 viii

1

8

c If Pr(A ∩ B ∩ C ) = Pr(A) × Pr(B) × Pr(C ) and A, B and C are ‘piecewise independent’ (that is, AB, AC and BC are all independent pairs of events), then A, B and C are independent.

7 a 957

b 0.746238319

≈

c 0.21669319

≈

d 0.20774357

≈ e Answers will vary. Make reference to

the percentage who developed asthma given the drug, compared with those not given the drug (20.7% compared with 24.3%).

f ξ

Asthma > 1Asthma = 1

Drug

143 176

84 97

62

395

g 176 h 62

i 0.24688

357≈

j 0.645176

273≈

Introductory probability - Hawker Maths 202011E Karnaugh maps and probability tables 11F Conditional...

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