Past exam of the year 1 of the computer science course of the University of Oxford / 2021 / Discrete mathematics, continuous mathematics, and probability / 8 / a / iii

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- Solution iii

Solution

Let

R_{1}, R_{2}

be the two red-card events. Sampling without replacement gives

P (R_{1} \cap R_{2}) = \frac{m}{N} \frac{m - 1}{N - 1} .

This differs from the product

m^{2} / N^{2}

, since

P (R_{1} \cap R_{2}) - P (R_{1}) P (R_{2}) = - \frac{mn}{N ^{2} ( N - 1 )} < 0.

Thus the events are not independent, for every positive

m, n

. A first red card reduces the remaining red fraction, a basic instance of how sampling without replacement need not be independent.

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