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.
Slide 9.3 - 3 Copyright © 2011 Pearson, Inc.
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
.
Slide 9.3 - 4 Copyright © 2011 Pearson, Inc.
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
Slide 9.3 - 5 Copyright © 2011 Pearson, Inc.
Example Rolling the Dice
Find the probability of rolling a sum divisible by 4 on a single roll of two fair dice.
Slide 9.3 - 6 Copyright © 2011 Pearson, Inc.
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.
Slide 9.3 - 7 Copyright © 2011 Pearson, Inc.
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.
Slide 9.3 - 8 Copyright © 2011 Pearson, Inc.
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.
Slide 9.3 - 9 Copyright © 2011 Pearson, Inc.
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
.
Slide 9.3 - 10 Copyright © 2011 Pearson, Inc.
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.
Slide 9.3 - 11 Copyright © 2011 Pearson, Inc.
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?
Slide 9.3 - 12 Copyright © 2011 Pearson, Inc.
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.
Slide 9.3 - 13 Copyright © 2011 Pearson, Inc.
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.
Slide 9.3 - 14 Copyright © 2011 Pearson, Inc.
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?
Slide 9.3 - 15 Copyright © 2011 Pearson, Inc.
Example Choosing Chocolates
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
.
Slide 9.3 - 16 Copyright © 2011 Pearson, Inc.
Conditional Probability Formula
If the event B depends on the event A, then
P(B | A) =P(A and B)P(A)
.
Slide 9.3 - 17 Copyright © 2011 Pearson, Inc.
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.
Slide 9.3 - 18 Copyright © 2011 Pearson, Inc.
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
Slide 9.3 - 19 Copyright © 2011 Pearson, Inc.
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?
Slide 9.3 - 20 Copyright © 2011 Pearson, Inc.
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?
P(15 successes) = 0.92( )15≈0.286
Slide 9.3 - 21 Copyright © 2011 Pearson, Inc.
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 exactly 10?
Slide 9.3 - 22 Copyright © 2011 Pearson, Inc.
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 exactly 10?
P(10 successes)=
15
10
⎛
⎝⎜⎞
⎠⎟0.92( )
100.08( )
5≈0.00427
Slide 9.3 - 23 Copyright © 2011 Pearson, Inc.
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