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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