Chapter 12 Probability © 2008 Pearson Addison-Wesley. All rights reserved.

15
Chapter 12 Probabilit y © 2008 Pearson Addison-Wesley. All rights reserved

Transcript of Chapter 12 Probability © 2008 Pearson Addison-Wesley. All rights reserved.

Page 1: Chapter 12 Probability © 2008 Pearson Addison-Wesley. All rights reserved.

Chapter 12

Probability

© 2008 Pearson Addison-Wesley.All rights reserved

Page 2: Chapter 12 Probability © 2008 Pearson Addison-Wesley. All rights reserved.

© 2008 Pearson Addison-Wesley. All rights reserved

12-4-2

Chapter 12: Probability

12.1 Basic Concepts

12.2 Events Involving “Not” and “Or”

12.3 Conditional Probability; Events Involving

“And”

12.4 Binomial Probability

12.5 Expected Value

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12-4-3

Chapter 1

Section 12-4Binomial Probability

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Binomial Probability

• Binomial Probability Distribution

• Binomial Probability Formula

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Binomial Probability Distribution

The spinner below is spun twice and we are interested in the number of times a 2 is obtained (assume each sector is equally likely).Think of outcome 2 as a “success” and outcomes 1 and 3 as “failures.” The sample space is 1

2

3

S = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}.

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Binomial Probability Distribution

When the outcomes of an experiment are divided into just two categories, success and failure, the associated probabilities are called “binomial.” Repeated trials of the experiment, where the probability of success remains constant throughout all repetitions, are also known as Bernoulli trials.

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Binomial Probability Distribution

If x denotes the number of 2s occurring on each pair of spins, then x is an example of a random variable. In S, the number of 2s is 0 in four cases, 1 in four cases, and 2 in one case. Because the table on the next slide includes all the possible values of x and their probabilities it is an example of a probability distribution. In this case, it is a binomial probability distribution.

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Probability Distribution for the Number of 2s in Two Spins

x P(x)

0

1

2

4

94

91

99

Sum 19

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Binomial Probability Formula

In general, letn = the number of repeated trials,p = the probability of success on any given trial,q = 1 – p = the probability of failure on any

given trial, and x = the number of successes that occur.

Note that p remains fixed throughout all n trials. This means that all trials are independent. In general, x, successes can be assigned among n repeated trials in

nCx different ways.

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Binomial Probability Formula

When n independent repeated trials occur, where p = probability of success and q = probability of failure

with p and q (where q = 1 – p) remaining constant throughout all n trials, the probability of exactly x successes is given by

!( ) .

!( )!x n x x n x

n xn

P x C p q p qx n x

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Example: Coin Tossing

Find the probability of obtaining exactly three heads in five tosses of a fair coin.

Solution

3 2

5 31 1 1 1 5

( 3) 10 .2 2 8 4 16

P x C

This is a binomial experiment with n = 5, p = 1/2, q = 1/2, and x = 3.

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Example: Rolling a Die

Solution

Find the probability of obtaining exactly two 3’s in six rolls of a fair die.

2 4

6 21 5 1 625

( 2) 15 .201.6 6 36 1296

P x C

This is a binomial experiment with n = 6, p = 1/6, q = 5/6, and x = 2.

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Example: Rolling a Die

Solution

Find the probability of obtaining less than two 3’s in six rolls of a fair die.

We have n = 6, p = 1/6, q = 5/6, and x < 2.

( 2) ( 0) ( 1)P x P x P x 0 6 1 5

6 0 6 11 5 1 5

6 6 6 6C C

.3349 .4019 .7368.

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Example: Baseball Hits

Solution

A baseball player has a well-established batting average of .250. In the next series he will bat 10 times. Find the probability that he will get more than two hits.

In this case n = 10, p = .250, q = .750, and x > 2.

( 2) 1 ( 2)P x P x

1 ( 0, 1, or 2)P x

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Example: Baseball Hits

Solution (continued)

1 ( 0, 1, or 2)P x

0 10 1 910 0 10 1

2 810 2

.25 .75 .25 .751

.25 .75

C C

C

1 .0563 .1877 .2816

.4744