Copyright © 2011 Pearson, Inc. 9.3 Probability. Copyright © 2011 Pearson, Inc. Slide 9.3 - 2 What you’ll learn about Sample Spaces and Probability Functions.

Copyright © 2011 Pearson, Inc.

9.3Probability

Slide 9.3 - 2 Copyright © 2011 Pearson, Inc.

What you’ll learn about

Sample Spaces and Probability Functions Determining Probabilities Venn Diagrams and Tree Diagrams Conditional Probability Binomial Distributions

… and whyEveryone should know how mathematical the “laws of chance” really are.


Probability of an Event(Equally Likely Outcomes)

If E is an event in a finite, nonempty sample space

S of equally likely outcomes, then the probability

of the event E is

P(E) =the number of outcomes in Ethe number of outcomes in S

.


Probability Distribution for the Sum of Two Fair Dice

Outcome Probability2 1/363 2/364 3/365 4/366 5/367 6/368 5/369 4/3610 3/3611 2/3612 1/36


Example Rolling the Dice

Find the probability of rolling a sum divisible by 4 on a single roll of two fair dice.


Example Rolling the Dice

Find the probability of rolling a sum divisible by 4 on a single roll of two fair dice.

The event E consists of the outcomes 4,8,12{ } .

To get the probability of E we add up theprobabilities of the outcomes in E:

P(E) =336

+536

+136

=936

=14.


Probability Function

A probability function is a function P that assigns

a real number to each outcome in a sample space S

subject to the following conditions:

1. 0 ≤P(O) ≤1;2. the sum of the probabilities of all outcomes in S is 1;3. P(∅) =0.


Probability of an Event(Outcomes not Equally Likely)

Let S be a finite, nonempty sample space in which

every outcome has a probability assigned to it by a

probability function P. If E is any event in S , the

probability of the event E is the sum of the

probabilities of all the outcomes contained in E.


Strategy for Determining Probabilities

1. Determine the sample space of all possible outcomes.

When possible, choose outcomes that are equally likely.

2. If the sample space has equally likely outcomes, the

probability of an event E is determined by counting:

P(E) =the number of outcomes in Ethe number of outcomes in S

.


Strategy for Determining Probabilities

3. If the sample space does not have equally likely

outcomes, determine the probability function.

(This is not always easy to do.) Check to be sure

that the conditions of a probability function are

satisfied. Then the probability of an event E is

determined by adding up the probabilities of all

the outcomes contained in E.


Example Choosing Chocolates

Dylan opens a box of a dozen chocolate cremes and offers three of them to Russell. Russell likes vanilla cremes the best, but all the chocolates look alike on the outside. If five of the twelve cremes are vanilla, what is the probability that all of Russell’s picks are vanilla?


Example Choosing ChocolatesIf five of the twelve cremes are vanilla, what is the probability that all of Russell’s 3 picks are vanilla?

The experiment in question is the selection of three

chocolates, without regard to order, from a box of 12.

There are 12

C3=220 outcomes of this experiment.

The event E consists of all possible combinations of 3 that can be chosen, without regard to order, from the5 vanilla cremes available. There are 5C3 =10 ways.

Therefore, P(E) =10 / 220 =1/ 22.


Multiplication Principle of Probability

Suppose an event A has probability p1 and an event B has probability p2 under the assumption that A occurs. Then the probability that both A and B occur is p1p2.



Dylan opens a box of a dozen chocolate cremes and offers three of them to Russell. Russell likes vanilla cremes the best, but all the chocolates look alike on the outside. If five of the twelve cremes are vanilla, what is the probability that all of Russell’s picks are vanilla?



The probability of picking a vanilla creme on the first

draw is 5/12. Under the assumption that a vanilla creme

was selected in the first draw, the probability of picking

a vanilla creme on the second draw is 4/11. Under the

assumption that a vanilla creme was selected in the first

and second draw, the probability of picking a vanilla

creme on the third draw is 3/10. By the Multiplication

Principle, the probability of picking

a vanilla creme on all three picks is 5

12⋅411

⋅310

=60

1320=

122

.


Conditional Probability Formula

If the event B depends on the event A, then

P(B | A) =P(A and B)P(A)

.


Binomial Distribution

Suppose an experiment consists of n-independent

repetitions of an experiment with two outcomes,

called "success" and "failure."

Let P(success) =p and P(failure) =q.(Note that q=1−p.)

Then the terms in the binomial expansion of (p+q)n

give the respective probabilities of exactlyn, n−1,..., 2, 1, 0 successes.


Binomial Distribution

Number of successes out of Probability

n independent repetitions

n pn

n−1 nn−1

⎛

⎝⎜⎞

⎠⎟pn−1q

r nr

⎛

⎝⎜⎞

⎠⎟prqn−r

1 nr

⎛

⎝⎜⎞

⎠⎟pqn−1

0 qn


Example Shooting Free Throws

Suppose Tommy makes 92% of his free throws. If he shoots 15 free throws, and if his chance of making each one is independent of the other shots, what is the probability that he makes all 15?



Suppose Tommy makes 92% of his free throws. If he shoots 15 free throws, and if his chance of making each one is independent of the other shots, what is the probability that he makes all 15?

P(15 successes) = 0.92( )15≈0.286



Suppose Tommy makes 92% of his free throws. If he shoots 15 free throws, and if his chance of making each one is independent of the other shots, what is the probability that he makes exactly 10?



Suppose Tommy makes 92% of his free throws. If he shoots 15 free throws, and if his chance of making each one is independent of the other shots, what is the probability that he makes exactly 10?

P(10 successes)=

15

10

⎛

⎝⎜⎞

⎠⎟0.92( )

100.08( )

5≈0.00427


Quick Review Solutions

How many outcomes are possible for the following

experiments.

1. Two coins are tossed. 4

2. Two different 6-sided dice are rolled. 36

3. Two chips are drawn simultaneously without

replacement from a jar with 8 chips. 28

4. Two different cards are drawn from a standard deck of 52.

1326

5. Evaluate without using a calculator. 4C

2

8C

2

3 14

Copyright © 2011 Pearson, Inc. 9.3 Probability. Copyright © 2011 Pearson, Inc. Slide 9.3 - 2 What you’ll learn about Sample Spaces and Probability Functions.

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