chapter 17 binomial and geometric probability models binomial and geometric random variables and...
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“2-outcome” situations are very common Heads/tails Democrat/Republican Male/Female Win/Loss Success/Failure Defective/NondefectiveTRANSCRIPT
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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