概率论英文课件：ch5_4 Hypergeometric Distribution

When n is small compared to N, the nature of the N items changes very little in each draw. Thus, the quantity k/N plays the role of the parameter p. As a result, the binomial distribution may be viewed as a large population edition of the hypergeometric distributions.
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Theorem 5.3 Mean and variance of the hypergeometric distribution h(x; N, n, k) are
= nk/N
,
N n k k n (1 ) N 1 N N
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Relationship to the Binomial Distribution
5.4 Hypergeometric Distribution

Hypergeometric Experiment A random sample of size n is selected without replacement from N items. k of the N items may be classified as successes and N – k are classified as failures.
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Example:
A state runs a lottery( 彩 票 ) in which six numbers are randomly selected from 40,without replacement. A player chooses six numbers before the state’s sample is selected. Let X be the numbers selected by the player that match those selected by the state. X has a Hypergeometric Distribution. Question: (a) What is the probability that six numbers chosen by player match all six numbers in state’s sample? N = 40, k = 6, x = 6 (b) What is the probability that five of the six numbers chosen by a player appear in the state’s sample? N = 40, k = 6, x = 5
4 6 1 9 0 3 3 0 3 7 10 10 10 10 3 3
= 54/100. So P(A’) = 46%.
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Hypergeometric Distribution Let X be the number of successes in a random sample of size n of a hypergeometric experiment, then X is called a hypergeometric random variable; the probability distribution of X is called the hypergeometric distribution, and its values P(X = x), x = 0, 1, …, n, are denoted by h(x; N, n, k). k N k h(x; N, n, k) = x n x , x = 0, 1, …, n. N n
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Example
A purchaser of electrical components buys them in lots of size 10. It is his policy to inspect 3 components randomly from a lot and to accept the lot only if all 3 are nondefective. If 30% of lots have 4 defective components and 70% have only one, what is the probability that a lot is rejected (or what percent of lots are rejected)? Solution: Let A denote the event that a lot is accepted. P(A) = P(A | lot has 4 defectives) 3/10 + P(A | lot has 1 defective)7/10 =