Exactly one nutty chocolate in two

Question

A box has 20 chocolates, of which 6 contain nuts and 14 do not. You take 2 chocolates at random. What is the probability that exactly 1 of the 2 contains nuts?

Accepted Answer

Taking 2 chocolates is an unordered pick, so count groups with combinations, where \binom{n}{k} counts ways to choose k from n. Total ways to choose 2 chocolates from 20: \binom{20}{2}=190. Favorable: exactly 1 with nuts and 1 without. Pick 1 of the 6 nutty and 1 of the 14 plain: \binom{6}{1}	imes\binom{14}{1}=6	imes 14=84. Probability: \frac{84}{190}=\frac{42}{95}\approx 0.442. Takeaway. For a two-item pick with "one of each kind," the count is simply (number of one type) times (number of the other), over all pairs. It is the hypergeometric formula at its smallest.

Exactly one nutty chocolate in two

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