Hypergeometric Distribution - Application and Example

Application and Example

The classical application of the hypergeometric distribution is sampling without replacement. Think of an urn with two types of marbles, black ones and white ones. Define drawing a white marble as a success and drawing a black marble as a failure (analogous to the binomial distribution). If the variable N describes the number of all marbles in the urn (see contingency table below) and m describes the number of white marbles, then N − m corresponds to the number of black marbles. In this example X is the random variable whose outcome is k, the number of white marbles actually drawn in the experiment. This situation is illustrated by the following contingency table:

	drawn	not drawn	total
white marbles	k	m − k	m
black marbles	n − k	N + k − n − m	N − m
total	n	N − n	N

Now, assume (for example) that there are 5 white and 45 black marbles in the urn. Standing next to the urn, you close your eyes and draw 10 marbles without replacement. What is the probability that exactly 4 of the 10 are white? Note that although we are looking at success/failure, the data are not accurately modeled by the binomial distribution, because the probability of success on each trial is not the same, as the size of the remaining population changes as we remove each marble.

This problem is summarized by the following contingency table:

	drawn	not drawn	total
white marbles	k = 4	m − k = 1	m = 5
black marbles	n − k = 6	N + k − n − m = 39	N − m = 45
total	n = 10	N − n = 40	N = 50

The probability of drawing exactly k white marbles can be calculated by the formula

Hence, in this example calculate

Intuitively we would expect it to be even more unlikely for all 5 marbles to be white.

$P(X=5) = f(5;50,5,10) = {{{5 \choose 5} {{45} \choose {5}}}\over {50 \choose 10}} = {1\cdot 1221759 \over 10272278170} = 0.0001189375\dots,$

As expected, the probability of drawing 5 white marbles is roughly 35 times less likely than that of drawing 4.

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