Seminar 7 MM150

Bashkim Zendeli

Chapter 7 PROBABILITY

WHAT YOU WILL LEARN

• Empirical probability and theoretical probability

• Compound probability, conditional probability, and binomial probability

• Odds against an event and odds in favor of an event

• Expected value• Tree diagrams

Definitions

An experiment is a controlled operation that yields a set of results.

The possible results of an experiment are called its outcomes.

An event is a subcollection of the outcomes of an experiment.

Definitions continued

Empirical probability is the relative frequency of occurrence of an event and is determined by actual observations of an experiment.

Theoretical probability is determined through a study of the possible outcomes that can occur for the given experiment.

Empirical Probability

Example: In 100 tosses of a fair die, 19 landed showing a 3. Find the empirical probability of the die landing showing a 3.

Let E be the event of the die landing showing a 3.

P(E)

number of times

event E has occurred

total number of times the

experiment has been performed

P(E)

19

1000.19

The Law of Large Numbers

The law of large numbers states that probability statements apply in practice to a large number of trials, not to a single trial. It is the relative frequency over the long run that is accurately predictable, not individual events or precise totals.

7.2

Theoretical Probability

Equally likely outcomes

If each outcome of an experiment has the same chance of occurring as any other outcome, they are said to be equally likely outcomes.

For equally likely outcomes, the probability of Event E may be calculated with the following formula.

P(E)

number of outcomes favorable to E

total number of possible outcomes

Example

A die is rolled. Find the probability of rolling a) a 2. b) an odd number. c) a number less than 4. d) an 8. e) a number less than 9.

Solutions: There are six equally likely outcomes: 1, 2, 3, 4, 5, and 6. a)

b) There are three ways an odd number can occur 1, 3 or 5.

c) Three numbers are less than 4.

P(2)

number of outcomes that will result in a 2

total number of possible outcomes

1

6

P(odd)

3

6

1

2

P(number less than 4)

3

6

1

2

d) There are no outcomes that will result in an 8.

e) All outcomes are less than 9. The event must occur and the probability is 1.

Solutions: There are six equally likely outcomes: 1, 2, 3, 4, 5, and 6. continued

P(number greater than 8)

0

60

Important Facts

The probability of an event that cannot occur is 0. The probability of an event that must occur is 1. Every probability is a number between 0 and 1

inclusive; that is, 0 P(E) 1. The sum of the probabilities of all possible

outcomes of an experiment is 1.

Example

A standard deck of cards is well shuffled. Find the probability that the card is selected.

a) a 10.

b) not a 10.

c) a heart.

d) an ace, one or 2.

e) diamond and spade.

f) a card greater than 4 and less than 7.

Example continued

a) a 10

There are four 10’s in a deck of 52 cards.

b) not a 10

P(10)

4

52

1

13

P(not a 10) 1 P(10)

11

13

12

13

Example continued

c) a heart

There are 13 hearts in the deck.

d) an ace, 2 or 3

There are 4 aces, 4 twos and 4 threes, or a total of 12 cards.

P(heart)

13

52

1

4 P(A, 1 or 2)

12

52

3

13

Example continued

d) diamond and spade

The word and means both events must occur. This is not possible.

e) a card greater than 4 and less than 7

The cards greater than 4 and less than 7 are 5’s, and 6’s.

P > 4 and < 7 P 5 or 6

8

52

2

13

P(diamond and spade)

0

520

7.3

Odds

Odds Against

Odds against event P event fails to occur

P event occurs

P failure P success

Example: Odds Against

Example: Find the odds against rolling a 5 on one roll of a die.

The odds against rolling a 5 are 5:1.

P 5 1

6 P fails to roll a 5 5

6

5616

5

66

1

5

1odds against rolling a 5

Odds in Favor

Odds in favor of event P event occurs

P event fails to occur

P success P failure

Example

Find the odds in favor of landing on blue in one spin of the spinner.

The odds in favor of spinning blue are 3:5.

P blue 3

8 P not blue 5

8

odds in favor

3858

3

88

5

3

5

Probability from Odds

Example: The odds against spinning a blue on a certain spinner are 4:3 Find the probability that

a) a blue is spun. b) a blue is not spun.

Solution

Since the odds are 4:3 the denominators must be 4 + 3 = 7.

The probabilities ratios are:

P blue 4

7

P not blue 3

7

7.4

Expected Value (Expectation)

Expected Value

The symbol P1 represents the probability that the first event will occur, and A1 represents the net amount won or lost if the first event occurs.

E P1A

1 P

2A

2 P

3A

3 ... P

nA

n

Example

Teresa is taking a multiple-choice test in which there are four possible answers for each question. The instructor indicated that she will be awarded 3 points for each correct answer and she will lose 1 point for each incorrect answer and no points will be awarded or subtracted for answers left blank. If Teresa does not know the correct answer to

a question, is it to her advantage or disadvantage to guess?

If she can eliminate one of the possible choices, is it to her advantage or disadvantage to guess at the answer?

Solution

Expected value if Teresa guesses.

P guesses correctly 1

4

E 1

43 3

4 1

3

4

3

40

P guesses incorrectly 3

4

Solution continued—eliminate a choice

P guesses correctly 1

3

E 1

33 2

3 1

12

3

1

3

P guesses incorrectly 2

3

Example: Winning a Prize

When Calvin Winters attends a tree farm event, he is given a free ticket for the $75 door prize. A total of 150 tickets will be given out. Determine his expectation of winning the door prize.

E 1

15075 149

1500

1

2

Example

When Calvin Winters attends a tree farm event, he is given the opportunity to purchase a ticket for the $75 door prize. The cost of the ticket is $3, and 150 tickets will be sold. Determine Calvin’s expectation if he purchases one ticket.

Solution

Calvin’s expectation is $2.48 when he purchases one ticket.

E 1

15075 149

150 3

75

150

447

150

372

150 2.48

Fair Price

Fair price = expected value + cost to play

Example

Suppose you are playing a game in which you spin the pointer shown in the figure, and you are awarded the amount shown under the pointer. If is costs $10 to play the game, determine

a) the expectation of the person who plays the game.

b) the fair price to play the game.

$10

$10

$2

$2

$20$15

Solution

$0

3/8

$10

$10$5$8Amount Won/Lost

1/81/83/8Probability

$20$15$2Amt. Shown on Wheel

E 3

8 $8 3

8$0 1

8$5 1

8$10

24

8 0

5

8

10

8

9

8 1.125 $1.13

Solution

Fair price = expectation + cost to play

= $1.13 + $10

= $8.87

Thus, the fair price is about $8.87.

7.5

Tree Diagrams

Counting Principle

If a first experiment can be performed in M distinct ways and a second experiment can be performed in N distinct ways, then the two experiments in that specific order can be performed in M • N distinct ways.

Definitions

Sample space: A list of all possible outcomes of an experiment.

Sample point: Each individual outcome in the sample space.

Tree diagrams are helpful in determining sample spaces.

Example

Two balls are to be selected without replacement from a bag that contains one purple, one blue, and one green ball.

a) Use the counting principle to determine the number of points in the sample space.

b) Construct a tree diagram and list the sample space.

c) Find the probability that one blue ball is selected.

d) Find the probability that a purple ball followed by a green ball is selected.

Solutions

a) 3 • 2 = 6 ways

b)

c)

d)

B

P

B

G

B

G

P

G

P

PB

PGBP

BGGP

GB

P blue 4

6

2

3

P Purple,Green P P,G 1

6

P(event happening at least once)

P

event happening

at least once

1 P

event does

not happen

7.6

Or and And Problems

Or Problems

P(A or B) = P(A) + P(B) P(A and B) Example: Each of the numbers 1, 2, 3, 4, 5, 6,

7, 8, 9, and 10 is written on a separate piece of paper. The 10 pieces of paper are then placed in a bowl and one is randomly selected. Find the probability that the piece of paper selected contains an even number or a number greater than 5.

Solution

P(A or B) = P(A) + P(B) P(A and B)

Thus, the probability of selecting an even number or a number greater than 5 is 7/10.

Peven or

greater than 5

P even P greater than 5 Peven and

greater than 5

5

10

5

10

3

10

7

10

Example

Each of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 is written on a separate piece of paper. The 10 pieces of paper are then placed in a bowl and one is randomly selected. Find the probability that the piece of paper selected contains a number less than 3 or a number greater than 7.

Solution

There are no numbers that are both less than 3 and greater than 7. Therefore,

P less than 3 2

10

P greater than 7 3

10

P

less than 3 or

greater than 7

2

10

3

10 0

5

10

1

2

Mutually Exclusive

Two events A and B are mutually exclusive if it is impossible for both events to occur simultaneously.

Example

One card is selected from a standard deck of playing cards. Determine the probability of the following events.

a) selecting a 3 or a jack

b) selecting a jack or a heart

c) selecting a picture card or a red card

d) selecting a red card or a black card

Solutions

a) 3 or a jack

b) jack or a heart

P 3 P jack 4

52

4

52

8

52

2

13

P jack P heart Pjack and

heart

4

52

13

52

1

52

16

52

4

13

Solutions continued

c) picture card or red card

d) red card or black card

P picture P red Ppicture &

red card

12

52

26

52

6

52

32

52

8

13

P red P black 26

52

26

52

52

521

And Problems

P(A and B) = P(A) • P(B) Example: Two cards are to be selected with

replacement from a deck of cards. Find the probability that two red cards will be selected.

P A P B P red P red

26

5226

52

1

21

2

1

4

Example

Two cards are to be selected without replacement from a deck of cards. Find the probability that two red cards will be selected.

red red

26 25

52 511 25 25

2 51 102

P A P B P P

Independent Events

Event A and Event B are independent events if the occurrence of either event in no way affects the probability of the occurrence of the other event.

Experiments done with replacement will result in independent events, and those done without replacement will result in dependent events.

Example

A package of 30 tulip bulbs contains 14 bulbs for red flowers, 10 for yellow flowers, and 6 for pink flowers. Three bulbs are randomly selected and planted. Find the probability of each of the following.a.All three bulbs will produce pink flowers.b.The first bulb selected will produce a red

flower, the second will produce a yellow flower and the third will produce a red flower.

c.None of the bulbs will produce a yellow flower.d.At least one will produce yellow flowers.

Solution

30 tulip bulbs, 14 bulbs for red flowers, 10 for yellow flowers, and 6 for pink flowers.

a. All three bulbs will produce pink flowers.

P 3 pink P pink 1 P pink 2 P pink 3

6

30

5

29

4

28

1

203

Solution

30 tulip bulbs, 14 bulbs for red flowers,0010 for yellow flowers, and 6 for pink flowers.

b. The first bulb selected will produce a red flower, the second will produce a yellow flower and the third will produce a red flower.

P red, yellow, red P red P yellow P red

14

3010

2913

28

13

174

Solution

30 tulip bulbs, 14 bulbs for red flowers,0010 for yellow flowers, and 6 for pink flowers.

c. None of the bulbs will produce a yellow flower.

Pnone

yellow

P

first not

yellow

P

second not

yellow

P

third not

yellow

20

3019

2918

28

57

203

Solution

30 tulip bulbs, 14 bulbs for red flowers, 10 for yellow flowers, and 6 for pink flowers.

d. At least one will produce yellow flowers.

P(at least one yellow) = 1 P(no yellow)

157

203

146

203

Questions***

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