slide 12-1 copyright © 2005 pearson education, inc. seventh edition and expanded seventh edition

1 Copyright © 2005 Pearson Education, Inc.

SEVENTH EDITION and EXPANDED SEVENTH EDITION

Copyright © 2005 Pearson Education, Inc.

Chapter 12

Probability


12.1

The Nature of Probability


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 outcome 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.

number of times event has occurred( )

total number of times the experiment has been performed

EP E

19( ) 0.19

100P E


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.


12.2

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.

number of outcomes favorable to ( )

total number of outcomes

EP E


Example

A die is rolled. Find the probability of rolling a) a 3. b) an odd number c) a number less than 4 d) a 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.

number of outcomes that will result in a 2 1(2)

total number of possible outcomes 6P

3 1(odd)

6 2P

3 1(number less than 4)

6 2P


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

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

e) All outcomes are less than 10. The event must occur and the probability is 1.

0(number greater than 8) 0

6P


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) a 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

4 1(10)

52 13P

(not a 10) 1 (10)

11

1312

13

P P


Example continued

c) a heart

There are 13 hearts in the deck.

d) an ace, 1 or 2

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

13 1(heart)

52 4P

12 3(A, 1 or 2)

52 13P


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.

0(diamond & spade) 0

52P 8 2

( )52 13

P E


12.3

Odds


Odds Against

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

(event fails to occur) ( )Odds against event =

(event occurs) ( )

P P failure

P P success

1(5)

6P

5(fails to roll a 5)

6P

56

16

5 6 5(odds)

6 1 1P The odds against

rolling a 5 are 5:1.


Odds in Favor

(event occurs)Odds in favor of event =

(event fails to occur)

( )

( )

P

P

P success

P failure


Example

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

3(blue)=

8P 5

(not blue)=8

P

38

58

3 8 3(odds in favor)

8 5 5P

The odds in favor of spinning blue are 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:

4

(blue)=7

P

3(not blue)=

7P


12.4

Expected Value (Expectation)


Expected Value

E = x1P(x1) + x2P(x2) + x3P(x3) +…+ xnP(xn)

or E = P(x1)x1 + P(x2)x2 + P(x3)x3 +…+ P(xn)xn

(order doesn’t matter)

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.


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.

1(guesses correctly)

4P

3(guesses incorrectly)

4P

1 3Teresa's expectation = (3) ( 1)

4 43 3

04 4


Solution continued—eliminate a choice

1(guesses correctly)

3P

2(guesses incorrectly)

3P

1 2Teresa's expectation = (3) ( 1)

3 32 1

13 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.

1 149Expectation = (75) (0)

150 1501

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 $2.50 when he purchases one ticket.

5.2150

375150

447

150

72

150

1493

150

172)(

xE


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$8Amt. Won/Lost

1/81/83/8Probability

$20$15$2Amt. Shown on Wheel

3 3 1 1Expectation = ( $8) ($0) ($5) ($10)

8 8 8 824 5 10

08 8 89

1.125 1.138


Solution

Fair price = expectation + cost to play

= $1.13 + $10

= $8.87

Thus, the fair price is about $8.87.


12.5

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 way, then the two experiments in that specific order can be performed in M • N ways. (Note: Order is important.)


Definitions

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

Sample point (simple event): Each individual outcome in the sample space.


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) P(purple then green =

1/6

Note: The probability of each simple event is B

P

B

G

B

G

P

G

P

PBPG

BPBG

GP

GB

4 2(blue)=

6 3P

.6

1

2

1

3

1


12.6

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.

even or even and (even) (greater 5)

greater than 5 greater than 5

5 5 3

10 10 107

10

P P P


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,

2(less than 3)

10P

3(greater than 7)

10P

2 3 5 10

10 10 10 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

4 4(3) ( jack)

52 528 2

52 13

P P

jack and 4 13 1( jack) (heart)

heart 52 52 52

16 4

52 13

P P P


Solutions continued

c) picture card or red card

d) red card or black card

26 26(red) (black)

52 5252

152

P P

picture & 12 26 6(picture) (red)

red card 52 52 52

32 8

52 13

P P P


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.

( ) ( ) ( ) ( )

26 26

52 521 1 1

2 2 4

P A P B P red P red


Example

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

102

25

2652

650

51

25

52

26

)()()(

redPredPredandredP


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. All three bulbs will produce pink flowers. The first bulb selected will produce a red flower, the

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

None of the bulbs will produce a yellow flower. 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.

All three bulbs will produce pink flowers.

3 pink (pink 1) (pink 2) (pink 3)

6 5 4=

30 29 281

203

P P P P


Solution


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

14 10 13=

30 29 2813

174

P P P P


Solution


None of the bulbs will produce a yellow flower.

first not second not third notnone yellow

yellow yellow yellow

20 19 18=

30 29 2857

203

P P P P


Solution


At least one will produce yellow flowers.

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

571

203146

203


12.7

Conditional Probability


Conditional Probability

In general, the probability of event E2 occurring, given that an event E1 has

happened is called conditional probability

and is written

P(E2|E1).


Example

Given a family of two children, and assuming that boys and girls are equally likely, find the probability that the family has a) two girls. b) two girls if you know that at least one of the

children is a girl. c) two girls given that the older child is a girl.


Solutions

a) two girls There are four possible outcomes BB, BG, GB,

and GG.

b) two girls if you know that at least one of the children is a girl

1(2 girls) =

4P

1(both girls|at least one is a girl) =

3P


Solutions continued

two girls given that the older child is a girl

S = {BB, BG, GB, GG}

Older is a Girl BG, GG

Two girls given older is a girl GG

1(both girls|older child is a girl) =

2P


Conditional Probability For any two events, E1, and E2,

)(

)(

)(

)(

)(

)(

)(

)()|(

1

21

1

21

1

1212

EventGivenP

EventsBothP

EP

EANDEP

En

EANDEn

En

EANDEnEEP

2

1

)(

)2(

)|2(

:

1

12

12

E

EE

EE

girlaisoldern

girlaisolderANDgirlsn

girlaisoldergirlsP

Example

S = {BB, BG, GB, GG}

Older is a Girl BG, GG

Two girls given older is a girl GG


Example

Use the results of the taste test given at a local mall. If one person from the sample is selected at random, find the probability the person selected

257127130Total

1327260Women

1255570Men

TotalPrefers

Wintergreen

Prefers Peppermint


Example continued

a) prefers peppermint

b) is a woman

130(peppermint)

257P

132(woman)

257P


Example continued

c) prefers peppermint, given that a woman is selected

d) is a man, given that the person prefers wintergreen

60 15(peppermint|woman)

132 33P

55(man|wintergreen)

127P


12.8

The Counting Principle and Permutations


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 ways.


Example

A password to logon to a computer system is to consist of 3 letters followed by 3 digits. Determine how many different passwords are possible if a) repetition of letters and digits is permitted b) repetition of letters and digits is not permitted c) the first letter must be a vowel (a, e, I, o, u) and

the first digit cannot be 0, and repetition of letters and digits is not permitted.


Solutions

repetition of letters and digits is permitted. There are 26 letters and 10 digits. We have 6

positions to fill.

L L L D D D

26 26 26 10 10 10

L L L D D D

17,576,000


Solution

repetition of letters and digits is not permitted.

L L L D D D

26 25 24 10 9 8

L L L D D D

11,232,000


Solution

the first letter must be a vowel (a, e, i, o, u) and the first digit cannot be 0, and repetition of letters and digits is not permitted.

L L L D D D

5 25 24 9 9 8

L L L D D D

1,944,000


Permutations

A permutation is any ordered arrangement of a given set of objects. (Order is important)

Number of Permutations The number of permutations of n distinct

items is n factorial, symbolized n!, where n! = n(n 1)(n 2) … (3)(2)(1)


Example

How many different ways can 6 stuffed animals be arranged in a line on a shelf?

6! = 6 • 5 • 4 • 3 • 2 • 1 = 720

The 6 stuffed animals can be arranged in 720 different ways.


Example

Consider the six numbers 1, 2, 3, 4, 5 and 6. In how many distinct ways can three numbers be selected and arranged if repetition is not allowed?

6 • 5 • 4 = 120

Thus, there are 120 different possible ordered arrangements, or permutations.


Permutation Formula

The number of permutations possible when r objects are selected from n objects is found by the permutation formula

!

!n r

nP

n r


Example

The swimming coach has 8 swimmers who can compete in the 100 m butterfly, he can only enter 3 swimmers in the event. In how many ways could he select the 3 swimmers?

8 3

8! 8!

8 3 ! 5!

8 7 6 5 4 3 2 1

P

5 4 3 2 1 336


Permutations of Duplicate Objects

The number of distinct permutations of n objects where n1 of the objects are identical, n2 of the objects are identical, …, nr of the objects are identical is found by the formula

1 2 3

!

! !... !

n

n n n


Example

In how many different ways can the letters of the word “CINCINNATI” be arranged?

Of the 10 letters, 2 are C’s, 3 are N’s, and 3 are I’s.

10! 1 60 9 8 7

3!3!2!

35 14 2 3 2 1 3 2 1 2 1

100,80050,400

2


12.9

Combinations


Combination

A combination is a distinct group (or set) of objects without regard to their arrangement.

The number of combinations possible when r objects are selected from n objects is found by

!

! !n r

nC

n r r


Example

A student must select 4 of 7 essay questions to be answered on a test. In how many ways can this selection be made?

There are 35 different ways that 4 of 7 questions can be selected.

7 4

7! 7!

(7 4)!4! 3!4!

7 6

C

5 4 3 2 1 3 2 1 4 3 2 1

35


Example

Toastline Bakery is testing 5 new wheat breads, 4 bran breads and 3 oat breads. If it plans to market 2 of the wheat breads, 2 of the bran breads and one of the oat breads, how many different combinations are possible?

5 2 4 2 3 1Bread choices

10 6 3

180

C C C


12.10

Solving Probability Problems by Using Combinations


Example

A club consists of 5 men and 6 women. Four members are to be selected at random to form a committee. What is the probability that the committee will consist of two women?

6 2

11 4

# of possible committees with two women(two women) =

total number of 4-member committees

15 1

330 22

P

C

C


Example

The Honey Bear is testing 10 new flavors of ice cream. They are testing 5 vanilla based, 3 chocolate based and 2 strawberry based ice creams. If we assume that each of the 10 flavors has the same chance of being selected and that 4 new flavors will be produced, find the probability that a) no chocolate flavors are selected. b) at least 1 chocolate is selected. c) 2 vanilla and 2 chocolate are selected.


Solution

5 vanilla, 3 chocolate, 2 strawberry

selecting 4 flavors

a)

b)

7 4

10 4

35 1(no chocolate)

210 6

CP

C

(at least 1 chocolate) 1 (no chocolate)

1 51

6 6

P P


Solution continued

5 2 3 2

10 4

(2 vanilla and 2 chocolate)

10 3 30 1

210 210 7

C CP

C


12.11

Binomial Probability Formula


To Use the Binomial Probability Formula

There are n repeated independent trials.

Each trial has two possible outcomes, success and failure.

For each trial, the probability of success (and failure) remains the same.


Binomial Probability Formula

The probability of obtaining exactly x successes, P(x), in n independent trials is given by

where p is the probability of success on a single trial and q is the probability of failure on a single trial.

( ) x n xn xP x C p q


Example

A basket contains 5 pens: one of each color red, blue, black, green and purple. Five pens are going to be selected with replacement from the basket. Find the probability that no blue pens are selected exactly 1 blue pen is selected exactly 2 blue pens are selected exactly 3 blue pens are selected


Solution

no blue pens exactly 1 blue pen

0 5 0

5 0

5

5

( )

1 4(0)

5 5

4(1)(1)

5

4 1024

5 3125

x n xn xP x C p q

P C

1 5 1

5 1

4

( )

1 4(0)

5 5

1 4(5)

5 5

1 256 2565

5 625 625

x n xn xP x C p q

P C


Solution continued

exactly 2 blue pens exactly 3 blue pens

3 5 3

5 3

3 2

( )

1 4(0)

5 5

1 4(5)

5 5

1 16 3210

125 25 625

x n xn xP x C p q

P C

2 5 2

5 2

2 3

( )

1 4(0)

5 5

1 4(10)

5 5

1 64 12810

25 125 625

x n xn xP x C p q

P C


Example

A manufacturer of lunch boxes knows that 0.7% of the lunch boxes are defective. Write the binomial probability formula that would be use

to determine the probability that exactly x our of n lunch boxes produced are defective.

Write the binomial probability formula that would be used to find the probability that exactly 4 lunch boxes out of 60 produced will be defective.


Solution

( )

(0.007) ((0.993)

x n xn x

x n xn x

P x C p q

C

4 56

60 4

( )

(0.007) ((0.993)

x n xn xP x C p q

C


Example

The probability that an egg selected at random has a weak shell and will crack is 0.2. Find the probability that none of 8 eggs selected at random has a weak

shell. at least one of 8 eggs selected has a weak shell.


Solution

p = 0.2 q = 1 0.2 = 0.8 We want 0 successes in 8 trials.

at least 1egg = 1 0.167772 = 0.832228

0 8 0

8 0

( )

(0.2) (0.8)

1(1)(0.167772)

0.167772

x n xn xP x C p q

C

slide 12-1 copyright © 2005 pearson education, inc. seventh edition and expanded seventh edition

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