hypergeometric distributions
DESCRIPTION
Hypergeometric Distributions. Remember, for rolling dice. uniform. binomial. geometric. Rolling a pair P(pair) = eventually. Rolling a 4 P(4) = 1/6. Rolling a 7 P(pair) = 6/36. A Hypergeometric distribution does not fit any of these models. It is close to the Binomial Model. - PowerPoint PPT PresentationTRANSCRIPT
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Hypergeometric Distributions
![Page 2: Hypergeometric Distributions](https://reader036.vdocuments.site/reader036/viewer/2022062500/56814f6a550346895dbd2164/html5/thumbnails/2.jpg)
Remember, for rolling dice
uniform
Rolling a 4
P(4) = 1/6
binomial
Rolling a 7
P(pair) = 6/36
geometricRolling a pairP(pair) = eventually
![Page 3: Hypergeometric Distributions](https://reader036.vdocuments.site/reader036/viewer/2022062500/56814f6a550346895dbd2164/html5/thumbnails/3.jpg)
A Hypergeometric distribution does not fit any of these
models.
It is close to the Binomial Model.
![Page 4: Hypergeometric Distributions](https://reader036.vdocuments.site/reader036/viewer/2022062500/56814f6a550346895dbd2164/html5/thumbnails/4.jpg)
HD are used when the probabilities of successive trials (or selections)
are dependent
Meeting the first condition determines the probability of the second.
The random variable (X) is the number of successful trials in an experiment.For efficiency, we will use combinations to help us count
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Jury SelectionA six member jury is needed.
There are 10 women and 6 men available.
Create a probability distribution for all the possible juries.
Identify the DRV X as the number of women on the jury.
Try a binomial model….
![Page 6: Hypergeometric Distributions](https://reader036.vdocuments.site/reader036/viewer/2022062500/56814f6a550346895dbd2164/html5/thumbnails/6.jpg)
Jury SelectionA six member jury is needed.
There are 10 women and 6 men available.
Create a probability distribution for all the possible juries.
X: a womanp = 10 / 16q = 6 /16
Suppose we select 4 women, so x = 4There is a breakdown when we model this…..not the same as dice….not all the probabilities stay the same
![Page 7: Hypergeometric Distributions](https://reader036.vdocuments.site/reader036/viewer/2022062500/56814f6a550346895dbd2164/html5/thumbnails/7.jpg)
Jury SelectionA six member jury is needed.
There are 10 women and 6 men available.
Create a probability distribution for all the possible juries.
X: a womanp = 10 / 16q = 6 /16
Suppose we select 4 women, so x = 4Our way around this is to chose the women in groups (like subsets, we have done this before….)
![Page 8: Hypergeometric Distributions](https://reader036.vdocuments.site/reader036/viewer/2022062500/56814f6a550346895dbd2164/html5/thumbnails/8.jpg)
Determine the PD for the number of women selected for a 6 member jury
from a group of10 women and 8 men
Let (the RV) X be the number of women selected.
Number of women (x) Probability P(x)
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P(x) = n(x)
n(S)
= women X men
Total number of
selections
= 10Cx X 8C6 - x
18C6
![Page 10: Hypergeometric Distributions](https://reader036.vdocuments.site/reader036/viewer/2022062500/56814f6a550346895dbd2164/html5/thumbnails/10.jpg)
Probability in a Hypergeometric Distribution
P(x) = aCx X n-aCr-x
nCrx: number of successful selections
r: size of the subset drawn (6)
a: total number of successful elements to draw from (10)
n: all selections (18)
![Page 11: Hypergeometric Distributions](https://reader036.vdocuments.site/reader036/viewer/2022062500/56814f6a550346895dbd2164/html5/thumbnails/11.jpg)
Expected ValueE(X) = r a
nWhat is the expected number of women on the committee?
r = 6
a = 10
n = 18E(X) = 6[10/18]
= 3.3
![Page 12: Hypergeometric Distributions](https://reader036.vdocuments.site/reader036/viewer/2022062500/56814f6a550346895dbd2164/html5/thumbnails/12.jpg)
Homework
Pg 404 1,3,7,11,12
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Remember
1. If you roll a die 4 times, what is the probability that
a) you roll a 3 twice?
b) you roll a 3 on the last roll?