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Marked chips variance

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A bag holds $10$ chips, of which $4$ are marked. You draw $3$ chips without replacement, with all subsets equally likely. What is the variance of the number of marked chips drawn?

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Hint

Use the hypergeometric variance: the binomial variance times a finite-population factor.

Solution

The count is hypergeometric, with variance $n \frac{( \frac{K}{N} ) ( 1 - \frac{K}{N} ) ( N - n )}{N - 1}$ .
Here this is $3 \times \frac{2}{5} \times \frac{3}{5} \times \frac{7}{9}$ .
The product is $\frac{18}{25} \times \frac{7}{9}$ , which equals $\frac{14}{25}$ .

Answer: $\frac{14}{25}$

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