CHAPTER 17 BINOMIAL AND CHAPTER 17 BINOMIAL AND GEOMETRIC PROBABILITY GEOMETRIC PROBABILITY MODELS 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
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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
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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.
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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