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Hypergeometric Distributions

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.

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Hypergeometric Distributions

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  1. Hypergeometric Distributions

  2. 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

  3. A Hypergeometric distribution does not fit any of these models.It is close to the Binomial Model.

  4. 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

  5. Jury Selection A 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….

  6. Jury Selection A six member jury is needed. There are 10 women and 6 men available. Create a probability distribution for all the possible juries. X: a woman p = 10 / 16 q = 6 /16 Suppose we select 4 women, so x = 4 There is a breakdown when we model this…..not the same as dice….not all the probabilities stay the same

  7. Jury Selection A six member jury is needed. There are 10 women and 6 men available. Create a probability distribution for all the possible juries. X: a woman p = 10 / 16 q = 6 /16 Suppose we select 4 women, so x = 4 Our way around this is to chose the women in groups (like subsets, we have done this before….)

  8. Determine the PD for the number of women selected for a 6 member jury from a group of10 women and 8 menLet (the RV) X be the number of women selected. Number of women (x) Probability P(x)

  9. = women X men Total number of selections = 10Cx X 8C6 - x 18C6 n(S) P(x) = n(x)

  10. Probability in a Hypergeometric Distribution P(x) = aCx X n-aCr-x nCr x: 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)

  11. Expected Value n E(X) = r a What is the expected number of women on the committee? r = 6 a = 10 n = 18 E(X) = 6[10/18] = 3.3

  12. Homework Pg 404 1,3,7,11,12

  13. Remember • If you roll a die 4 times, what is the probability that • you roll a 3 twice? b) you roll a 3 on the last roll?

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