CHAPTER 17 BINOMIAL AND GEOMETRIC PROBABILITY MODELS Binomial and Geometric Random Variables and Their Probability Distributions.

CHAPTER 17 BINOMIAL AND CHAPTER 17 BINOMIAL AND GEOMETRIC PROBABILITY GEOMETRIC PROBABILITY

MODELSMODELSBinomial and Geometric Random

Variables and Their 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

Classic binomial experiment: Classic binomial experiment: tossing atossing acoin a pre-specified number coin a pre-specified number of timesof times

Toss a coin 10 timesResult of each toss: head or tail (designate

one of the outcomes as a success, the other as a failure; makes no difference)

P(head) and P(tail) are the same on each toss

trials are independent◦ if you obtained 9 heads in a row, P(head) and

P(tail) on toss 10 are same as P(head) and P(tail) on any other toss (not due for a tail on toss 10)

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.

ExamplesExamplesa. Yes; n=10; success=“major

repairs within 3 months”; p=.05b. No; n not specified in advancec. No; p changesd. Yes; n=1500; success=“chip is

defective”; p=.10

Binomial Probability Binomial Probability DistributionDistribution

0 0

trials, success probability on each trialprobability 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

ExampleExampleA production line produces motor

housings, 5% of which have cosmetic defects. A quality control manager randomly selects 4 housings from the production line. Let x=the number of housings that have a cosmetic defect. Tabulate the probability distribution for x.

SolutionSolution(i) D=defective, G=goodoutcome x P(outcome)GGGG 0 (.95)(.95)(.95)(.95)DGGG 1 (.05)(.95)(.95)(.95)GDGG 1 (.95)(.05)(.95)(.95) : : :DDDD 4 (.05)4

SolutionSolution

0 44 0

1 34 1

2 24 2

3 14 3

44 4

( ) is a binomial random variable

( ) , 0,1, 2, ,4, .05 ( .95)

(0) (.05) (.95) .815

(1) (.05) (.95) .171475

(2) (.05) (.95) .01354

(3) (.05) (.95) .00048

(4) (.05) (.9

x n xn x

ii x

p x C p q x nn p q

p C

p C

p C

p C

p C

05) .00000625

SolutionSolution

x 0 1 2 3 4p(x) .815

.171475 .01354 .00048 .00000625

Example (cont.)Example (cont.)x 0 1 2 3 4p(x) .815

.171475 .01354 .00048 .00000625

What is the probability that at least 2 of the housings will have a cosmetic defect?

P(x p(2)+p(3)+p(4)=.01402625

Example (cont.)Example (cont.)

What is the probability that at most 1 housing will not have a cosmetic defect? (at most 1 failure=at least 3 successes)

P(x )=p(3) + p(4) = .00048+.00000625 = .00048625

x 0 1 2 3 4p(x) .815 .171475 .01354 .00048 .00000625

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 = 2σ = √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

Recall from Chap. 16 Recall from Chap. 16 Random Variables: Hardee’s Random Variables: Hardee’s

vs. The Colonelvs. The Colonel

Hardee’s vs The ColonelHardee’s vs The ColonelOut of 100 taste-testers, 63

preferred Hardee’s fried chicken, 37 preferred KFC

Evidence that Hardee’s is better? A landslide?

What if there is no difference in the chicken? (p=1/2, flip a fair coin)

Is 63 heads out of 100 tosses that unusual?

Use binomial rv to analyzeUse binomial rv to analyzen=100 taste testersx=# who prefer Hardees chickenp=probability a taste tester

chooses HardeesIf p=.5, P(x 63) = .0061 (since

the probability is so small, p is probably NOT .5; p is probably greater than .5, that is, Hardee’s chicken is probably better).

Recall from Chap. 16 Recall from Chap. 16 Random Variables: Random Variables:

Mothers Identify Mothers Identify NewbornsNewborns

After spending 1 hour with their newborns, blindfolded and nose-covered mothers were asked to choose their child from 3 sleeping babies by feeling the backs of the babies’ hands

22 of 32 women (69%) selected their own newborn

“far better than 33% one would expect…”Is it possible the mothers are guessing?Can we quantify “far better”?

Use binomial rv to Use binomial rv to analyzeanalyze

n=32 mothersx=# who correctly identify their own babyp= probability a mother chooses her own

babyIf p=.33, P(x 22)=.000044 (since the

probability is so small, p is probably NOT .33; p is probably greater than .33, that is, mothers are probably not guessing.

Geometric Random Geometric Random VariablesVariables

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Geometric Random Geometric Random VariablesVariablesGeometric Probability Distributions

Through 2/25/2014 NC State’s free-throw percentage is 65.1 (315th of 351 in Div. 1). In the 2/26/2014 game with UNC what is the probability that the first missed free-throw by the ‘Pack occurs on the 5th attempt?

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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 independentThe binomial rv counts the number of

successes in the n trials

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The Geometric ModelThe Geometric ModelA geometric random variable

counts the number of trials until the first success is observed.

A geometric random variable is completely specified by one parameter, p, the probability of success, and is denoted Geom(p).

Unlike a binomial random variable, the number of trials is not fixed

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The Geometric Model The Geometric Model (cont.)(cont.)

Geometric probability model for Bernoulli trials: Geom(p)

p = probability of successq = 1 – p = probability of failureX = # of trials until the first success

occursp(x) = P(X = x) = qx-1p, x = 1, 2,

3, 4,…1( )E Xp

2

qp

37

The Geometric Model (cont.)The Geometric Model (cont.)The 10% condition: the trials must

be independent. If that assumption is violated, it is still okay to proceed as long as the sample is smaller than 10% of the population.

Example: 3% of 33,000 NCSU students are from New Jersey. If NCSU students are selected 1 at a time, what is the probability that the first student from New Jersey is the 15th student selected?

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ExampleExampleThe American Red Cross says that about 11% of

the U.S. population has Type B blood. A blood drive is being held in your area.

1. How many blood donors should the American Red Cross expect to collect from until it gets the first donor with Type B blood?

Success=donor has Type B bloodX=number of donors until get first donor with

Type B blood

1 1.11; ( ) 9.09.11

p E Xp

39

Example (cont.)Example (cont.)The American Red Cross says that about 11%

of the U.S. population has Type B blood. A blood drive is being held in your area.

2. What is the probability that the fourth blood donor is the first donor with Type B blood?

4 1 4 1 3(4) (.89) (.11) .89 .11 .0775p q p

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Example (cont.)Example (cont.)The American Red Cross says that about 11%

of the U.S. population has Type B blood. A blood drive is being held in your area.

3. What is the probability that the first Type B blood donor is among the first four people in line?

0 1 2 3

.11;have to find(1) (2) (3) (4)

(.89 .11) (.89 .11) (.89 .11) (.89 .11).11 .0979 .087 .078 .3729

pp p p p

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Geometric Probability Distributionp = 0.1

0

0.02

0.04

0.06

0.08

0.1

0.12

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

0 2

1 3

(1) .9 .1 .1 (3) .9 .1 .081(2) .9 .1 .09 (4) .9 .1 .0729

1 1( ) 10.1

p pp p

E Xp

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Geometric Probability Distributionp = 0.25

0

0.05

0.1

0.15

0.2

0.25

0.3

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

0 2

1 3

(1) .75 .25 .25 (3) .75 .25 .141

(2) .75 .25 .1875 (4) .75 .25 .10551 1( ) 4

.25

p p

p p

E Xp

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

1. The expected number of attempts until she makes her first 3-point shot is what value?

2. What is the probability that the first 3-point shot she makes occurs on her 3rd attempt?

2(3) .75 .25 .141p

1 1( ) 4.25

E Xp

Question from earlier slide Question from earlier slide Through 2/25/2014 NC State’s free-

throw percentage was 65.1%. In the 2/26/2014 game with UNC what is the probability that the first missed free-throw by the ‘Pack occurs on the 5th attempt?

“Success” = missed free throwSuccess p = 1 - .651 = .349p(5) = .6514 .349 = .0627

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