Transcript 5.4 Hypergeometric 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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
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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
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



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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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
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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
=
 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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