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12 ProbabilityAssessment statements5.2 Conceptsoftrial,outcome,equallylikelyoutcomes,samplespace(U )and

event.TheprobabilityofaneventAasP(A) 5 n(A)/n(U ).ThecomplementaryeventsasAandA9(notA);P(A) 1 P(A9) 5 1.

UseofVenndiagrams,treediagramsandtablesofoutcomestosolveproblems.

5.3 Combinedevents,theformula:P(A⋃B) 5 P(A) 1 P(B) 2 P(A⋂B).P(A⋂B) 5 0formutuallyexclusiveevents.

5.4 Conditionalprobability;thedefinition:P(A∣B) 5 P(A⋂B)/P(B).Independentevents;thedefinition:P(A∣B) 5 P(A) 5 P(A∣B9).UseofBayes’theoremforamaximumofthreeevents.

Now that you have learned to describe a data set in Chapter 11, how can you use sample data to draw conclusions about the populations from which you drew your samples?

The techniques we use in drawing conclusions are part of what we call inferential statistics, which is a part of one of the HL options. Inferential statistics uses probability as one of its tools. To use this tool properly, you must first understand how it works. This chapter will introduce you to the language and basic tools of probability.

The variables we discussed in Chapter 11 can now be redefined as random variables, whose values depend on the chance selection of the elements in the sample. Using probability as a tool, you will be able to create probability distributions that serve as models for random variables. You can then describe these using a mean and a standard deviation as you did in Chapter 11.

12.1 Randomness

Probability is the study of randomness and uncertainty.

The reasoning in statistics rests on asking, ‘How often would this method give a correct answer if I used it very many times?’ When we produce data by random sampling or by experiments, the laws of probability enable us to answer the question, ‘What would happen if we did this many times?’

Introduction

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What does ‘random’ mean? In ordinary speech, we use ‘random’ to denote things that are unpredictable. Events that are random are not perfectly predictable, but they have long-term regularities that we can describe and quantify using probability. In contrast, haphazard events do not necessarily have long-term regularities. Take, for example, the tossing of an unbiased coin and observing the number of heads that appear. This is random behaviour.

When you throw the coin, there are only two outcomes, heads or tails. Figure 12.1 shows the results of the first 50 tosses of an experiment that tossed the coin 5000 times. Two sets of trials are shown. The red graph shows the result of the first trial: the first toss was a head followed by a tail, making the proportion of heads to be 0.5. The third toss was also a tail, so the proportion of heads is 0.33, then 0.25. On the other hand, the other set of trials, shown in green, starts with a series of tails, then a head, which raises the proportion to 0.2, etc.

The proportion of heads is quite variable at first. However, in the long run, and as the number of tosses increases, the proportion of heads stabilizes around 0.5. We say that 0.5 is the probability of a head.

It is important that you know that the proportion of heads in a small number of tosses can be far from the probability. Probability describes only what happens in the long run. How a fair coin lands when it is tossed is an example of a random event. One cannot predict perfectly whether the coin will land heads or tails. However, in repeated tosses, the fraction of times the coin lands heads will tend to settle down to a limit of 50%. The outcome of an individual toss is not perfectly predictable, but the long-term average behaviour is predictable. Thus, it is reasonable to consider the outcome of tossing a fair coin to be random.

Imagine the following scenario:

I drive every day to school. Shortly before school, there is a traffic light. It appears that it is always red when I get there. I collected data over the course of one year (180 school days) and considered the green light to be a ‘success’. Here is a partial table of the collected data.

Please distinguish between random and haphazard (chaos). At first glance they might seem to be the same because neither of their outcomes can be anticipated with certainty.

Prop

ortio

n of

hea

ds

10

0.2

0.4

0.6

0.8

1

0.1

0.3

0.5

0.7

0.9

11 21 31 41 51Number of throws

Figure 12.1

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Day 1 2 3 4 5 6 7 …

Light red green red green red red red …

Percentage green 0 50 33.3 50 40 33.3 28.6 …

The first day it was red, so the proportion of success is 0% (0 out of 1); the second day it was green, so the frequency is now 50% (1 out of 2); the third day it was red again, so 33.3% (1 out of 3), and so on. As we collect more data, the new measurement becomes a smaller and smaller fraction of the accumulated frequency, so, in the long run, the graph settles to the real chance of finding it green, which in this case is about 30%. The graph is shown below.

Actually, if you run a simulation for a longer period, you can see that it really stabilizes around 30%. See graph below.

You have to observe here that the randomness in the experiment is not in the traffic light itself, as it is controlled by a timer. In fact, if the system works well, it may turn green at the same time every day. The randomness of the event is the time I arrive at the traffic light.

The French Count Buffon (1707–1788) tossed a coin 4040 times and received 2048 heads, i.e. a proportion of 50.69%. Also, the English statistician Karl Pearson (1857–1936) tossed a coin 24 000 times and received 12 012 heads, a 50.05% proportion for heads.

Count Buffon

Perc

enta

ge s

ucce

ss

10

11 21 31 41 51 61 71 81 91 101 111 121 131 141 151 161 171

20

40

60

10

30

50

Number of times

Perc

enta

ge o

f suc

cess

10

20

40

60

10

30

50

101 201 301 401 501Number of times

519

If we ask for the probability of finding the traffic light green in the above example, our answer will be about 30%. We base our answer on knowing that, in the long run, the fraction of time that the traffic light was green is 30%. We could also say that the long-run relative frequency of the green light settles down to about 30%.

12.2 Basic definitions

Data is obtained by observing either uncontrolled events in nature or controlled situations in a laboratory. We use the term experiment to describe either method of data collection.

An experiment is the process by which an observation (or measurement) is obtained. A random (chance) experiment is an experiment where there is uncertainty concerning which of two or more possible outcomes will result.

Tossing a coin, rolling a die and observing the number on the top surface, counting cars at a traffic light when it turns green, measuring daily rainfall in a certain area, etc. are a few experiments in this sense of the word.

A description of a random phenomenon in the language of mathematics is called a probability model. For example, when we toss a coin, we cannot know the outcome in advance. What do we know? We are willing to say that the outcome will be either heads or tails. Because the coin appears to be balanced, we believe that each of these outcomes has probability 0.50. This description of coin tossing has two parts:• A list of possible outcomes.• A probability for each outcome.

This two-part description is the starting point for a probability model. We will begin by describing the outcomes of a random phenomenon and learn how to assign probabilities to the outcomes by using one of the definitions of probability.

The sample space S of a random experiment (or phenomenon) is the set of all possible outcomes.

For example, for one toss of a coin, the sample space is

S 5 {heads, tails}, or simply {h, t}

Example 1

Toss a coin twice (or two coins once) and record the results. What is the sample space?

Solution

S 5 {hh, ht, th, tt}

The notation for sample space could also be U (IB notation) or any other letter.

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Probability12

Example 2

Toss a coin twice (or two coins once) and count the number of heads showing. What is the sample space?

Solution

S 5 {0, 1, 2}

A simple event is the outcome we observe in a single repetition (trial) of the experiment.

For example, an experiment is throwing a die and observing the number that appears on the top face. The simple events in this experiment are {1}, {2}, {3}, {4}, {5} and {6}. Of course, the set of all these simple events is the sample space of the experiment.

We are now ready to define an event. There are several ways of looking at it, which in essence are all the same.

An event is an outcome or a set of outcomes of a random experiment.

With this understanding, we can also look at the event as a subset of the sample space or as a collection of simple events.

Example 3

When rolling a standard six-sided die, what are the sets of event A ‘observe an odd number’, and event B ‘observe a number less than 5’.

Solution

Event A is the set {1, 3, 5}. Event B is the set {1, 2, 3, 4}.

Sometimes it helps to visualize an experiment using some tools of set theory. Basically, there are several similarities between the ideas of set theory and probability, and it is very helpful when we see the connection. A simple but powerful diagram is the Venn diagram. The diagram shows the outcomes of the die rolling experiment.

4

2B

A

35

6

1

In general, in this book, we will use a rectangle to represent the sample space and closed curves to represent events, as shown in Example 3.

To understand the definitions more clearly, let’s look at the following additional example.

Set theory provides a foundation for all of mathematics. The language of probability is much the same as the language of set theory. Logical statements can be interpreted as statements about sets. This will enable us later to introduce a method of understanding how to set up probability problems that we need to tackle.

521

Example 4

Suppose we choose one card at random from a deck of 52 playing cards, what is the sample space S?

Solution

S 5 {A♣, 2♣, … K♣, A♦, 2♦, … K♦, A♥, 2♥, … K♥, A♠, 2♠, … K♠}

Some events of interest:

K 5 event of king 5{K♣, K♦, K♥, K♠}

H 5 event of heart 5 {A♥, 2♥, … K♥}

J 5 event of jack or better

5 {J♣, J♦, J♥, J♠, Q♣, Q♦, Q♥, Q♠, K♣, K♦, K♥, K♠, A♣, A♦, A♥, A♠}

Q 5 event of queen 5 {Q♣, Q♦, Q♥, Q♠}

Example 5

Toss a coin three times and record the results. Show the event ‘observing two heads’ as a Venn diagram.

Solution

The sample space is made up of 8 possible outcomes such as hhh, hht, tht, etc.

A

hhh

ttt

tth

hthhht

thh

htt

thtS

Observing exactly two heads is an event with three elements: {hht, hth, thh}.

K♣

Q♣

K

Q K♦

Q♦

K♠

J♠J

S

J♣

A♣ Q♠K♥

Q♥A♥

J♥

H

2♥

4♥

Some useful set theory resultsSet operations have a number of properties, which are basic consequences of the definitions.

Some examples are:

A B 5 B A(A9)9 5 AA S 5 AA S 5 SA A9 5 [A A9 5 S

S is the sample space and [ is the empty set.Two mainly valuable properties are known as De Morgan’s laws, which state that:(A B)9 5 A9 B9

(A B)9 5 A9 B9

And finallyA (B C ) 5 (A  B) (A C )A (B C ) 5 (A B) (A C )

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Probability12

Tree diagrams, tables and grids

In an experiment to check the blood types of patients, the experiment can be thought of as a two-stage experiment: first we identify the type of the blood and then we classify the Rh factor 1 or 2.

The simple events in this experiment can be counted using another tool, the tree diagram, which is extremely powerful and helpful in solving probability problems.

A

A�

A�B�

B�

AB�

AB�O�

O�

Blood type Rh factor Outcome

�

�

�

�

�

�

�

�

B

AB

O

Our sample space in this experiment is the set {A1, A2, B1, B2, AB1, AB2, O1, O2} as we can read from the last column.

This data can also be arranged in a probability table:

Blood type

Rh factor A B AB O

Positive A1 B1 AB1 O1

Negative A2 B2 AB2 O2

Or using a 2-dimensional grid

A B AB O

O�

�

�

B�

as shown right:

Example 6�

Two tetrahedral dice, one blue and one yellow, are rolled. List the elements of the following events:

T 5 {3 appears on at least one die}

B 5 {the blue die is a 3}

S 5 {sum of the dice is a six} 1

4

23

523

Solution

11

2

Blue

die

Yellow die

4

2 3 4

3

B

T

S

T 5 {(1, 3), (2, 3), (3, 3), (4, 3), (3, 4), (3, 2), (3, 1)}

B 5 {(1, 3), (2, 3), (3, 3), (4, 3)}

S 5 {(2, 4), (3, 3), (4, 2)}

1 In a large school, a student is selected at random. Give a reasonable sample space for answers to each of the following questions:

a) Are you left-handed or right-handed?

b) What is your height in centimetres?

c) How many minutes did you study last night?

2 We throw a coin and a standard six-sided die and we record the number and the face that appear in that order. For example, (5, h) represents a 5 on the die and a head on the coin. Find the sample space.

3 We draw cards from a deck of 52 playing cards.

a) List the sample space if we draw one card at a time.

b) List the sample space if we draw two cards at a time.

c) How many outcomes do you have in each of the experiments above?

4 Tim carried out an experiment where he tossed 20 coins together and observed the number of heads showing. He repeated this experiment 10 times and got the following results:

11, 9, 10, 8, 13, 9, 6, 7, 10, 11

a) Use Tim’s data to get the probability of obtaining a head.

b) He tossed the 20 coins for the 11th time. How many heads should he expect to get?

c) He tossed the coins 1000 times. How many heads should he expect to see?

5 In the game ‘Dungeons and Dragons’, a four-sided die with sides marked with 1, 2, 3 and 4 spots is used. The intelligence of the player is determined by rolling the die twice and adding 1 to the sum of the spots.

a) What is the sample space for rolling the die twice? (Record the spots on the 1st and 2nd throws.)

b) What is the sample space for the intelligence of the player?

Exercise 12.1 and 12.2

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Probability12

6 A box contains three balls, blue, green and yellow. You run an experiment where you draw a ball, look at its colour and then replace it and draw a second ball.

a) What is the sample space of this experiment?

b) What is the event of drawing yellow first?

c) What is the event of drawing the same colour twice?

7 Repeat the same exercise as in question 6 above, without replacing the first ball.

8 Nick flips a coin three times and each time he notes whether it is heads or tails.

a) What is the sample space of this experiment?

b) What is the event that heads occur more often than tails?

9 Franz lives in Vienna. He and his family decided that their next vacation will be to either Italy or Hungary. If they go to Italy, they can fly, drive or take the train. If they go to Hungary, they will drive or take a boat. Letting the outcome of the experiment be the location of their vacation and their mode of travel, list all the points in the sample space. Also list the sample space of the event ‘fly to destination.’

10 A hospital codes patients according to whether they have health insurance or no insurance, and according to their condition. The condition of the patient is rated as good (g), fair (f ), serious (s), or critical (c). The clerk at the front desk marks 0, for non-insured patients, and 1 for insured, and uses one of the letters for the condition. So, (1, c) means an insured patient with critical condition.

a) List the sample space of this experiment.

b) What is the event ‘not insured, in serious or critical condition’?

c) What is the event ‘patient in good or fair condition’?

d) What is the event ‘patient has insurance’?

11 A social study investigates people for different characteristics. One part of the study classifies people according to gender (G1 5 female, G2 5 male), drinking habits (K1 5 abstain, K2 5 drinks occasionally, K3 5 drinks frequently), and marital status (M1 5 married, M2 5 single, M3 5 divorced, M4 5 widowed).

a) List the elements of an appropriate sample space for observing a person in this study.

b) Define the following events: A 5 the person is a male, B 5 the person drinks, and C 5 the person is single List the elements of each A, B and C.

c) Interpret the following events in the context of this situation:

A B; A C; C9; A B C; A9 B.

12 Cars leaving the highway can take a right turn (R), left turn (L), or go straight (S). You are collecting data on traffic patterns at this intersection and you group your observations by taking four cars at a time every 5 minutes.

a) List a few outcomes in your sample space U. How many are there?

b) List the outcomes in the event that all cars go in the same direction.

c) List the outcomes that only two cars turn right.

d) List the outcomes that only two cars go in the same direction.

13 You are collecting data on traffic at an intersection for cars leaving a highway. Your task is to collect information about the size of the vehicle: truck (T), bus (B), car (C). You also have to record whether the driver has the safety belt on (SY) or no safety belt (SN), as well as whether the headlights are on (O) or off (F).

a) List the outcomes of your sample space, U.

b) List the outcomes of the event SY that the driver has the safety belt on.

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12.3 Probability assignments

There are a few theories of probability that assign meaning to statements like ‘the probability that A occurs is p%’. In this book, we will primarily examine only the relative frequency theory. In essence, we will follow the idea that probability is ‘the long-run proportion of repetitions on which an event occurs’. This allows us to ‘merge’ two concepts into one.

• Equally likely outcomes In the theory of equally likely outcomes, probability has to do with symmetries and the indistinguishability of outcomes. If a given experiment or trial has n possible outcomes among which there is no preference, they are equally likely. The probability of each outcome

is then 100% _____ n or 1 __ n . For example, if a coin is balanced well, there is no

reason for it to land heads in preference to tails when it is tossed, so,

c) List the outcomes of the event C that the vehicle you are recording is a car.

d) List the outcomes of the event in C SY, C9, and C SY.

14 Many electric systems use a built in ‘back-up’ system so that the equipment using the system will work even if some parts fail. Such a system is given in the diagram below.

Two parts of this system are installed ‘in parallel’, so that the system will work if at least one of them works. If we code a working system by 1 and a failing system by 0, then one of the outcomes would be (1, 0, 1), which means parts A and C work while B failed.

a) List the outcomes of your sample space, U.

b) List the outcomes of the event X that exactly 2 of the parts work.

c) List the outcomes of the event Y that at least 2 of the parts work?

d) List the outcomes of the event Z that the system functions.

e) List the outcomes of the events: Z9, X Z, X Z, Y Z, and Y Z.

15 Your school library has 5 copies of George Polya’s How To Solve It book. Copies 1 and 2 are first-edition, and copies 3, 4 and 5 are second edition. You are searching for a first-edition book, and you will stop when you find a copy. For example, if you find copy 2 immediately, then the outcome is 2. Outcome 542 represents the outcome that a first edition was found on the third attempt.

a) List the outcomes of your sample space, U.

b) List the outcomes of the event A that two books must be searched.

c) List the outcomes of the event B that at least two books must be searched.

d) List the outcomes of the event C that copy 1 is found.

B

C

A

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accordingly, the probability that the coin lands heads is equal to the

probability that it lands tails, and both are 100% _____ 2

5 50%. Similarly, if

a die is fair, the chance that when it is rolled it lands with the side with 1

on top is the same as the chance that it shows 2, 3, 4, 5 or 6: 100% _____ 6

or 1 __ 6

.

In the theory of equally likely outcomes, probabilities are between 0% and 100%. If an event consists of more than one possible outcome, the chance of the event is the number of ways it can occur divided by the total number of things that could occur. For example, the chance that a die lands showing an even number on top is the number of ways it could land showing an even number (2, 4 or 6) divided by the total number of things that could occur (6, namely showing 1, 2, 3, 4, 5 or 6).

• Frequency theoryIn the frequency theory, probability is the limit of the relative frequency with which an event occurs in repeated trials. Relative frequencies are always between 0% and 100%. According to the frequency theory of probability, ‘the probability that A occurs is p%’ means that if you repeat the experiment over and over again, independently and under essentially identical conditions, the percentage of the time that A occurs will converge to p. For example, to say that the chance a coin lands heads is 50% means that if you toss the coin over and over again, independently, the ratio of the number of times the coin lands heads to the total number of tosses approaches a limiting value of 50%, as the number of tosses grows. Because the ratio of heads to tosses is always between 0% and 100%, when the probability exists it must be between 0% and 100%.

Using Venn diagrams and the ‘equally likely’ concept, we can say that the probability of any event is the number of elements in an event A divided by the total number of elements in the sample space S. This is equivalent to

saying: P(A) 5 n(A)

____ n(S )

, where n(A) represents the number of outcomes in A

and n(S) represents the total number of outcomes. So, in Example 5, the probability of observing exactly two heads is: P(2 heads) 5 3 _ 8 .

Probability rulesRegardless of which theory we subscribe to, the probability rules apply.

Rule 1

Any probability is a number between 0 and 1, i.e. the probability P(A) of any event A satisfies 0 < P(A) < 1. If the probability of any event is 0, the event never occurs. Likewise, if the probability is 1, it always occurs. In rolling a standard die, it is impossible to get the number 9, so P(9) 5 0. Also, the probability of observing any integer between 1 and 6, inclusive, is 1.

Rule 2

All possible outcomes together must have a probability of 1, i.e. the probability of the sample space S is 1: P(S ) 5 1. Informally, this is sometimes called the ‘something has to happen rule’.

In all theories, probability is on a scale of 0% to 100%. ‘Probability’ and ‘chance’ are synonymous.

No matter how little a chance you think an event has, there is no such thing as negative probability.

No matter how large a chance you think an event has, there is no such thing as a probability larger than 1!

A B

S

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Rule 3If two events have no outcomes in common, the probability that one or the other occurs is the sum of their individual probabilities. Two events that have no outcomes in common, and hence can never occur together, are called disjoint events or mutually exclusive events.

P(A or B) 5 P(A) 1 P(B)

This is the addition rule for mutually exclusive events.

For example, in tossing three coins, the events of getting exactly two heads or exactly two tails are disjoint, and hence the probability of getting exactly

two heads or two tails is 3 _ 8 1 3 _ 8 5 6 _ 8 5 3 _ 4 .

Bhtt

tht

thh

ttt

Ahhh

hth

hht

tth

S

Additionally, we can always add the probabilities of outcomes because they are always disjoint. A trial cannot come out in two different ways at the same time. This will give you a way to check whether the probabilities you assigned are legitimate.

Rule 4Suppose that the probability that you receive a 7 on your IB exam is 0.2, then the probability of not receiving a 7 on the exam is 0.8. The event that contains the outcomes not in A is called the complement of A, and is denoted by A9.

A A�

P(A9) 5 1 2 P(A), or P(A) 5 1 2 P(A9).

Example 7

Data for traffic violations was collected in a certain country and a summary is given below:

Age group 18–20 years 21–29 years 30–39 years Over 40 years

Probability 0.06 0.47 0.29 0.18

What is the probability that the offender is a) in the youngest age group, b) between 21 and 40, and c) younger than 40?

You have to be careful with these rules. By the ‘something has to happen’ rule, the total of the probabilities of all possible outcomes must be 1. This is so because they are disjoint, and their sum covers all the elements of the sample space. Suppose someone reports the following probabilities for students in your high school (4 years). If the probability that a grade 1, 2, 3 or 4 student is chosen at random from the high school is 0.24, 0.24, 0.25 and 0.19 respectively, with no other possibilities, you should know immediately that there is something wrong. These probabilities add up to 0.92. Similarly, if someone claims that these probabilities are 0.24, 0.28, 0.25, 0.26 respectively, there is also something wrong. These probabilities add up to 1.03, which is more than 1.

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Solution

Each probability is between 0 and 1, and the probabilities add up to 1. Therefore, this is a legitimate assignment of probabilities.

a) The probability that the offender is in the youngest group is 6%.

b) The probability that the driver is in the group 21 to 39 years is 0.47 1 0.29 5 0.76.

c) The probability that a driver is younger than 40 years is 1 2 0.18 5 0.82.

Example 8

It is a striking fact that when people create codes for their cellphones, the first digits follow distributions very similar to the following one:

First digit 0 1 2 3 4 5 6 7 8 9

Probability 0.009 0.300 0.174 0.122 0.096 0.078 0.067 0.058 0.051 0.045

a) Find the probabilities of the following three events:

A 5 {first digit is 1}

B 5 {first digit is more than 5}

C 5 {first digit is an odd number}

b) Find the probability that the first digit is (i)1 or greater than 5, (ii) not 1, and (iii) an odd number or a number larger than 5.

Solution

a) From the table:P(A) 5 0.300

P(B) 5 P(6) 1 P(7) 1 P(8) 1 P(9)5 0.067 1 0.058 1 0.051 1 0.0455 0.221

P(C ) 5 P(1) 1 P(3) 1 P(5) 1 P(7) 1P(9)5 0.300 1 0.122 1 0.078 1 0.058 1 0.0455 0.603

b) (i) Since A and B are mutually exclusive, by the addition rule, the probability that the first digit is 1 or greater than 5 is

P(A or B ) 5 0.300 1 0. 221 5 0.521.

(ii) Using the complement rule, the probability that the first digit is not 1 is

P(A9) 5 1 2 P(A) 5 1 2 0.300 5 0.700.

(iii) The probability that the first digit is an odd number or a number larger than 5:

P(B or C ) 5 P(1) 1 P(3) 1 P(5) 1 P(6) 1 P(7) 1 P(8) 1 P(9)5 0.300 1 0.122 1 0.078 1 0.067 1 0.058 1 0.051

1 0.0455 0.721

Hint: Notice here that P(B or C ) is not the sum of P(B ) and P(C ) because B and C are not disjoint.

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Equally likely outcomesIn some cases we are able to assume that individual outcomes are equally likely because of some balance in the experiment. Tossing a balanced coin renders heads or tails equally likely, with each having a probability of 50%, and rolling a standard balanced die gives the numbers from 1 to 6 as equally likely, with each having a probability of 1 _ 6 .

Suppose in Example 8 we consider all the digits to be equally likely to happen, then our table would be:

First digit 0 1 2 3 4 5 6 7 8 9

Probability 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1

P(A) 5 0.1

P(B) 5 P(6) 1 P(7) 1 P(8) 1 P(9) 5 4 3 0.1 5 0.4

P(C ) 5 P(1) 1 P(3) 1 P(5) 1 P(7) 1 P(9) 5 5 3 0.1 5 0.5

Also, by the complement rule, the probability that the first digit is not 1 is

P(A9) 5 1 2 P(A) 5 1 2 0.1 5 0.9.

2-dimensional grids are also very helpful tools that are used to visualize 2-stage or sequential probability models. For example, consider rolling a normal unbiased cubical die twice. Here are some events and how to use the grid in calculating their probabilities:

11

2

Seco

nd ro

ll

First roll

6

2 3 4 5 6

3

4

5

P (same number) � 636

P (at least one six) � 1136

If we are interested in the probability that at least one roll shows a 6, we count the points on the column corresponding to 6 on the first roll and the points on the row corresponding to 6 on the second roll observing naturally that the point in the corner should not be counted twice.

If we are interested in the number showing on both rolls to be the same, then we count the points on the diagonal as shown.

Finally, if we are interested in the probability that the first roll shows a number larger than the second roll, then we pick the points below the diagonal.

Hence, P(first number . second number) 5 15 __ 36 .

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Geometric probabilitySome cases give rise to interpreting events as areas in the plane. Take for example shooting at a circular target at random. What is the probability of hitting the central part?

The probability of hitting the central part is given by

P 5 p (  R __

4 )

2 ______

pR2 5 1 ___ 16

.

Example 9�

Lydia and Rania agreed to meet at the ‘museum quarter’ between 12:00 and 13:00. The first person to arrive will wait 15 minutes. If the second person does not show up, the first person will leave and they meet afterwards. Assuming that their arrivals are at random, what is the probability that they meet?

Solution

If Lydia arrives x minutes after 12:00 and Rania arrives y minutes after 12:00, then the condition for them to meet is |x 2 y | < 15, and x < 60, y < 60.

Geometrically, the outcomes of their ‘encounter’ region is given in the shaded region in the diagram right.

The area for each triangle is 1 _ 2 bh 5 1 _ 2 (45)2, so, the shaded area is 602 2 452.

The probability they meet is

therefore 602 2 452 ________

602 5 7 ___ 16

.

Probability calculation for equally likely outcomes using counting principles

In an experiment where all outcomes are equally likely, the theoretical probability of an event A is given by

P(A) 5 n(A)

____ n(S )

where n(A) is the number of outcomes that make up the event A, and n(S ) is the total number of outcomes in the sample space.

The new ideas we want to discuss here involve the calculation of n(A) and n(S). Such calculations will involve what you learned in Chapter 4 about counting principles.

R

R4

0

15

60

(45)2

15 60

12

(45)212

531

Example 10�

In a group of 18 students, eight are females. What is the probability of choosing five students

a) with all girls?

b) with three girls and two boys?

c) with at least one boy?

Solution

The total number of outcomes is the number of ways we can choose 5 out of the 18 students. So

n(S ) 5 (  18 5 ) 5 8586.

a) This event will require that we pick our group from among the 8 girls. So,

n(A) 5 (  8 5 ) 5 56 ⇒ P(A) 5 56 ____ 8568

5 0.0065.

b) This event will require that we pick three out of the 8 girls, and at the same time, we pick 2 out of the 10 boys. So, using the multiplication principle,

n(B) 5 (  8 3 ) (  10 2 ) 5 56.45 5 2520 ⇒ P(B) 5 2520 ____ 8568

5 0.294.

Note: Did you observe that (  8 5 ) 5 (  8 3 ) ? Why?

c) This event can be approached in two ways:

• To have at least 1 boy means that we can have 1, 2, 3, 4 or 5 boys. These are mutually exclusive, so the probability in question is the sum

P(C ) 5 (  10 1 ) (  8 4 ) 1 (  10 2 ) (  8 3 ) 1 (  10 3 ) (  8 2 ) 1 (  10 4 ) (  8 1 ) 1 (  10 5 ) (  8 0 )

(  18 5 ) 5 8512 ____

8568 5 0.9935

• To recognize that at least 1 boy is the complement of no boys at all, i.e. 0 boys or all 5 girls.

P(C ) 5 1 2 P(A) 5 1 2 0.0065 5 0.9935.

Example 11�

A deck of playing cards has 52 cards. In a game, the player is given five cards. Find the probability of the player having

a) three cards of one denomination and two cards of another (three 7s and two Js for example).

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Probability12

This game can be played at two stages, First, the player is given five cards, and then he/she can decide to exchange some of the cards. (The cards exchanged are discarded and not returned to the deck!)

A player was given the following hand: Q♠, Q♦, Q♥, 4♣, 9♠. She decided to change the last two cards. Find the probability of the player having

b) three cards of one denomination and two cards of another

c) four queens.

Solution

a) The sample space consists of all possible 5-card hands that can be given out:

n(S ) 5 (  52 5 ) 5 2 598 960

Call the event of interest A.

As there are 13 denominations in the deck of cards then there are 13 choices for the first required denomination. Once a denomination is chosen, say 9,

then there are (  4 3 ) ways of choosing 3 cards out of the four. Using the

multiplication rule, there are 13 (  4 3 ) ways of choosing 3 cards of the

first denomination. We are now left with 12 possible denominations

for the second one, each can give us (  4 2 ) ways of getting two of the

cards, and hence using the multiplication rule, there are 12 (  4 2 ) ways

of choosing the cards for the second denomination. Again using the

multiplication rule we will have [ 13 (  4 3 ) ] [ 12 (  4 2 ) ] 5 3744 ways of

choosing the first and second denominations.

The requested probability is then

P(A) 5 3744 _______ 2 598 960

0.001 44.

b) Since we have 3 queens, then we need only look for 2 cards of a different denomination. Now, there are only 47 cards left in the deck

because we had 5 already. So the sample space has n(S ) 5 (  47 2 ) 5 1081

ways of getting the rest of the 5 cards. The other cards could be two 4’s, two 9’s or two of the rest of the 10 denominations.

We have (  3 2 ) 5 3 ways of getting two 4’s since 4♣ is already discarded.

We also have (  3 2 ) 5 3 ways to get two 9’s. Or, for each of the other 10

denominations (no Q, no 4 and no 9), we have (  4 2 ) 5 6 different ways

of getting two of them, i.e. we have 10 (  4 2 ) 5 60

different ways of

getting two cards of the same denomination other than Q, 4 or 9.

So, the total number of ways of getting two cards of the same denomination is 3 1 3 1 60 5 66 ways.

So, the required probability is P(A) 5 n(A)

____ n(S )

5 66 ____ 1081

0.0611.

533

c) To have 4 Q’s we only have to look for one, and there is only one way of getting the missing Q♣. That leaves us with one card to be chosen from the 46 cards left. 46 ways!

Therefore, P(A) 5 46 ____ 1081

0.0426.

1 In a simple experiment, chips with integers 1–20 inclusive were placed in a box and one chip was picked at random.

a) What is the probability that the number drawn is a multiple of 3?

b) What is the probability that the number drawn is not a multiple of 4?

2 The probability an event A happens is 0.37.

a) What is the probability that it does not happen?

b) What is the probability that it may or may not happen?

3 You are playing with an ordinary deck of 52 cards by drawing cards at random and looking at them.

a) Find the probability that the card you draw is (i) the ace of hearts (ii) the ace of hearts or any spade (iii) an ace or any heart (iv) not a face card.

b) Now you draw the ten of diamonds, put it on the table and draw a second card. What is the probability that the second card is (i) the ace of hearts? (ii) not a face card?

c) Now you draw the ten of diamonds, return it to the deck and draw a second card. What is the probability that the second card is (i) the ace of hearts? (ii) not a face card?

4 On Monday morning, my class wanted to know how many hours students spent studying on Sunday night. They stopped schoolmates at random as they arrived and asked each, ‘How many hours did you study last night?’ Here are the answers of the sample they chose on Monday, 14 January, 2008.

Number of hours 0 1 2 3 4 5

Number of students 4 12 8 3 2 1

a) Find the probability that a student spent less than three hours studying Sunday night.

b) Find the probability that a student studied for two or three hours.

c) Find the probability that a student studied less than six hours.

5 We throw a coin and a standard six-sided die and we record the number and the face that appear. Find

a) the probability of having a number larger than 3

b) the probability that we receive a head and a 6.

6 A die is constructed in a way that a 1 has the chance to occur twice as often as any other number.

a) Find the probability that a 5 appears.

b) Find the probability an odd number will occur.

Exercise 12.3

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Probability12

7 You are given two fair dice to roll in an experiment.

a) Your first task is to report the numbers you observe. (i) What is the sample space of your experiment? (ii) What is the probability that the two numbers are the same? (iii) What is the probability that the two numbers differ by 2? (iv) What is the probability that the two numbers are not the same?

b) In a second stage, your task is to report the sum of the numbers that appear. (i) What is the probability that the sum is 1? (ii) What is the probability that the sum is 9? (iii) What is the probability that the sum is 8? (iv) What is the probability that the sum is 13?

8 The blood types of people can be one of four types: O, A, B or AB. The distribution of people with these types differs from one group of people to another. Here are the distributions of blood types for randomly chosen people in the US, China and Russia.

Blood type

CountryO A B AB

US 0.43 0.41 0.12 ?

China 0.36 0.27 0.26 0.11

Russia 0.39 0.34 ? 0.09

a) What is the probability of type AB in the US?

b) Dirk lives in the US and has type B blood. What is the probability that a randomly chosen US citizen can donate blood to Dirk? (Type B can only receive from O and B.)

c) What is the probability of randomly choosing an American and a Chinese (independently) with type O blood?

d) What is the probability of randomly choosing an American, a Chinese and a Russian (independently) with type O blood?

e) What is the probability of randomly choosing an American, a Chinese and a Russian (independently) with the same blood type?

9 In each of the following situations, state whether or not the given assignment of probabilities to individual outcomes is legitimate. Give reasons for your answer.

a) A die is loaded such that the probability of each face is according to the following assignment (x is the number of spots on the upper face and P(x) is its probability.)

x 1 2 3 4 5 6

P(x) 0 1 _ 6 1 _ 3 1 _ 3 1 _ 6 0

b) A student at your school categorized in terms of gender and whether they are diploma candidates or not.P(female, diploma candidate) 5 0.57, P(female, not a diploma candidate) 5 0.23,P(male, diploma candidate) 5 0.43, P(male, not a diploma candidate) 5 0.18.

c) Draw a card from a deck of 52 cards (x is the suit of the card and P(x) is its probability).

x Hearts Spades Diamonds Clubs

P(x) 12 ___ 52 15 ___ 52 12 ___ 52 13 ___ 52

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10 In Switzerland, there are three ‘official’ mother tongues, German, French and Italian. You choose a Swiss at random and ask, ‘What is your mother tongue?’ Here is the distribution of responses:

Language German French Italian Other

Probability 0.58 0.24 0.12 ?

a) What is the probability that a Swiss person’s mother tongue is not one of the official ones?

b) What is the probability that a Swiss person’s mother tongue is not German?c) What is the probability that you choose two Swiss independent of each other

and they both have German mother tongue? d) What is the probability that you choose two Swiss independent of each other

and they both have the same mother tongue?

11 The majority of email messages are now ‘spam.’ Choose a spam email message at random. Here is the distribution of topics:

Topic Adult Financial Health Leisure Products Scams

Probability 0.165 0.142 0.075 0.081 0.209 0.145

a) What is the probability of choosing a spam message that does not concern these topics?

Parents are usually concerned with spam messages with ‘adult’ content and scams.b) What is the probability that a randomly chosen spam email falls into one of

the other categories?

12 Consider n to be a positive integer. Let f  (x) 5 (  n  x 1 1 )

_______ (  n  x   ) 

    , where x is also a positive

integer. Determine the values of x (in terms of n) for which f  (x) < 1.

13 Determine n in each of the following cases:

a) (  n  2 ) 5 190 b) (  n  4 ) 5 (  n  8 )

14 An experiment involves rolling a pair of dice, 1 white and 1 red, and recording the numbers that come up. Find the probability

a) that the sum is greater than 8

b) that a number greater than 4 appears on the white die

c) that at most a total of 5 appears.

15 Three books are picked from a shelf containing 5 novels, 3 science books and a thesaurus. What is the probability that

a) the thesaurus is selected?

b) two novels and a science book are selected?

16 Five cards are chosen at random from a deck of 52 cards. Find the probability that the set contains

a) 3 kings

b) 4 hearts and 1 diamond.

17 A class consists of 10 girls and 12 boys. A team of 6 members is to be chosen at random. What is the probability that the team contains

a) one boy?

b) more boys than girls?