Chapter 17 Probability Models Binomial Probability Models Poisson Probability Models

Chapter 17Chapter 17Probability ModelsProbability Models

Binomial Probability ModelsPoisson Probability Models

Binomial Random Binomial Random VariablesVariables

Binomial Probability Distributions

Binomial Random Binomial Random VariablesVariables

Through 2/25/2014 NC State’s free-throw percentage is 65.1% (315th out 351 in Div. 1).

If in the 2/26/2014 game with UNC, NCSU shoots 11 free-throws, what is the probability that:NCSU makes exactly 8 free-throws?NCSU makes at most 8 free throws?NCSU makes at least 8 free-throws?

““2-outcome” situations are 2-outcome” situations are very commonvery common

Heads/tailsDemocrat/RepublicanMale/FemaleWin/LossSuccess/FailureDefective/Nondefective

Probability Model for this Probability Model for this Common SituationCommon Situation

Common characteristics◦repeated “trials”◦2 outcomes on each trial

Leads to Binomial Experiment

Binomial ExperimentsBinomial Experimentsn identical trials

◦n specified in advance2 outcomes on each trial

◦usually referred to as “success” and “failure”

p “success” probability; q=1-p “failure” probability; remain constant from trial to trial

trials are independent

Binomial Random VariableBinomial Random VariableThe binomial random variable X is the number of “successes” in the n trials

Notation: X has a B(n, p) distribution, where n is the number of trials and p is the success probability on each trial.

Binomial Probability Binomial Probability DistributionDistribution

0 0

trials, success probability on each trial

probability distribution:

( ) , 0,1,2, ,

( ) ( )

( ) (

x n xn x

n nn x n xx

x x

n p

p x C p q x n

E x xp x x p q np

Var x E x npq

P(x) = • px • qn-xn ! (n – x )!x!

Number of outcomes with

exactly x successes

among n trials

Rationale for the Binomial Probability Formula

P(x) = • px • qn-xn ! (n – x )!x!

Number of outcomes with

exactly x successes

among n trials

Probability of x successes

among n trials for any one

particular order

Binomial Probability Formula

Graph of Graph of p(x)p(x); ; xx binomial binomial n=10 p=.5; p(0)+p(1)+ n=10 p=.5; p(0)+p(1)+ …… +p(10)=1+p(10)=1

Think of p(x) as the areaof rectangle above x

p(5)=.246 is the areaof the rectangle above 5

The sum of all theareas is 1

Binomial Probability Histogram: n=100, p=.5

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

Binomial Probability Histogram: n=100, p=.95

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

0.11

0.12

0.13

0.14

0.15

0.16

0.17

0.18

70 72 74 76 78 80 82 84 86 88 90 92 94 96 98 100

Binomial Distribution Binomial Distribution Example: Tennis First Example: Tennis First ServesServes

A tennis player makes 60% of herfirst serves and attempts 4 first-serves.Assume the outcomes of first-serves

are independent.What is the probability that exactly 3

of her first serves are successful?

3 14 3

4;success=successful first serve; .60

(3) (.60) (.40) 4(.216)(.40) .3456

n p

p C

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Binomial Distribution ExampleBinomial Distribution Example

• Shanille O’Keal is a WNBA player who makes 25% of her 3-point attempts.

• Assume the outcomes of 3-point shots are independent.1. If Shanille attempts 7 3-point shots in a game, what is

the expected number of successful 3-point attempts?2. Shanille’s cousin Shaquille O’Neal makes 10% of his 3-

point attempts. If they each take 12 3-point shots, who has the smaller probability of making 4 or fewer 3-point shots?

Shaq: 12, .10; ( 4) .996n p P X ( ) 7(.25) 1.75E X np

Shanille: 12, .25; ( 4) .842n p P X Shanille has the smaller probability.

Using binomial tables; Using binomial tables; n=20, p=.3n=20, p=.3

P(x 5) = .4164P(x > 8) = 1- P(x 8)=

1- .8867=.1133P(x < 9) = ?P(x 10) = ?P(3 x 7)=P(x 7) - P(x 2)

.7723 - .0355 = .7368

9, 10, 11, … , 20

8, 7, 6, … , 0 =P(x 8)

1- P(x 9) = 1- .9520

Binomial n = 20, p = .3 Binomial n = 20, p = .3 (cont.)(cont.)P(2 < x 9) = P(x 9) - P(x 2)

= .9520 - .0355 = .9165P(x = 8) = P(x 8) - P(x 7)

= .8867 - .7723 = .1144

Color blindness

The frequency of color blindness (dyschromatopsia) in the Caucasian American male population is estimated to be about 8%. We take a random sample of size 25 from this population.

We can model this situation with a B(n = 25, p = 0.08) distribution.

What is the probability that five individuals or fewer in the sample are color blind?

Use Excel’s “=BINOMDIST(number_s,trials,probability_s,cumulative)”

P(x ≤ 5) = BINOMDIST(5, 25, .08, 1) = 0.9877

What is the probability that more than five will be color blind?

P(x > 5) = 1 P(x ≤ 5) =1 0.9877 = 0.0123

What is the probability that exactly five will be color blind?

P(x = 5) = BINOMDIST(5, 25, .08, 0) = 0.0329

0%

5%

10%

15%

20%

25%

30%

0 2 4 6 8

10

12

14

16

18

20

22

24

Number of color blind individuals (x)

P(X

= x

)

Probability distribution and histogram for

the number of color blind individuals

among 25 Caucasian males.

x P(X = x) P(X <= x) 0 12.44% 12.44%1 27.04% 39.47%2 28.21% 67.68%3 18.81% 86.49%4 9.00% 95.49%5 3.29% 98.77%6 0.95% 99.72%7 0.23% 99.95%8 0.04% 99.99%9 0.01% 100.00%

10 0.00% 100.00%11 0.00% 100.00%12 0.00% 100.00%13 0.00% 100.00%14 0.00% 100.00%15 0.00% 100.00%16 0.00% 100.00%17 0.00% 100.00%18 0.00% 100.00%19 0.00% 100.00%20 0.00% 100.00%21 0.00% 100.00%22 0.00% 100.00%23 0.00% 100.00%24 0.00% 100.00%25 0.00% 100.00%

B(n = 25, p = 0.08)

What are the mean and standard deviation of the

count of color blind individuals in the SRS of 25

Caucasian American males?

µ = np = 25*0.08 = 2

σ = √np(1 p) = √(25*0.08*0.92) = 1.36

p = .08n = 10

p = .08n = 75

µ = 10*0.08 = 0.8 µ = 75*0.08 = 6

σ = √(10*0.08*0.92) = 0.86 σ = √(75*0.08*0.92) = 2.35

What if we take an SRS of size 10? Of size 75?

Recall Free-throw Recall Free-throw questionquestion

Through 2/25/14 NC State’s free-throw percentage was 65.1% (315th in Div. 1).

If in the 2/26/14 game with UNC, NCSU shoots 11 free-throws, what is the probability that:

1. NCSU makes exactly 8 free-throws?

2. NCSU makes at most 8 free throws?

3. NCSU makes at least 8 free-throws?

1. n=11; X=# of made free-throws; p=.651

p(8)= 11C8 (.651)8(.349)3

=.2262. P(x ≤ 8)=.798

3. P(x ≥ 8)=1-P(x ≤7)=1-.5717 = .4283

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Poisson Probability Poisson Probability ModelsModels

The Poisson experiment typically models situations where rare events occur over a fixed amount of time or within a specified region

Examples◦The number of cellphone calls per

minute arriving at a cellphone tower.◦The number of customers per hour

using an ATM◦The number of concussions per game

experienced by the participants.

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◦Properties of the Poisson experiment1) The number of successes (events) that

occur in a certain time interval is independent of the number of successes that occur in another time interval.

2) The probability of a success in a certain time interval is

the same for all time intervals of the same size, proportional to the length of the interval.

3) The probability that two or more successes will occur in an interval approaches zero as the interval becomes smaller.

Poisson Poisson ExperimentExperiment

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The Poisson Random Variable◦The Poisson random variable X is the

number of successes that occur during a given time interval or in a specific region

Probability Distribution of the Poisson Random Variable.

( ) 0,1,2...!

( ) ( )

where 0is the average number of occurences

in an interval of time or in a specified region.

xep x x

xE X V X

Poisson Prob Dist Poisson Prob Dist =1=1

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0 1 2 3 4 5

1 0

1 1

1 2

1( 0) (0) .3679

0!

1( 1) (1) .3679

1!

1( 2) (2) .1839

2!

eP X p

eP X p

eP X p

Poisson Prob Dist Poisson Prob Dist =5=5

00.020.040.060.080.1

0.120.140.160.180.2

0 1 2 3 4 5 6 7 8 9 10

0067

0337

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Example ◦Cars arrive at a tollbooth at a rate of

360 cars per hour.◦What is the probability that only two

cars will arrive during a specified one-minute period?

The probability distribution of arriving cars for any one-minute period is Poisson with = 360/60 = 6 cars per minute. Let X denote the number of arrivals during a one-minute period. 0446.

!26e

)2X(P26

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◦Example (cont.)◦What is the probability that at least

four cars will arrive during a one-minute period?

◦P(X>=4) = 1 - P(X<=3) = 1 - .151 = .849