Direct combinatorial approach:\(Probability=\frac{# \ of \ favorable \ outcomes}{total \ # \ of \ outcomes}\):
Favorable outcomes: \(C^1_1*C^1_4\) --> \(C^1_1\) choosing one rotten out of 1 and \(C^1_4\) choosing any for the second apple;
Total # of outcomes: \(C^2_5\), choosing 2 apples out of 5;
\(P=\frac{C^1_1*C^1_4}{C^2_5}=\frac{2}{5}\).
Reverse combinatorial approach:Let's count the probability of the opposite event and subtract this value from 1. The opposit event will be if out of 2 apples both are good, so \(P=1-\frac{C^2_4}{C^2_5}=\frac{2}{5}\).
Direct probability approach:\(P=\frac{1}{5}*\frac{4}{4}+\frac{4}{5}*\frac{1}{4}=\frac{2}{5}\): probability of {first apple rotten, second apple good} plus the probability of {first apple good, second apple rotten}.
Reverse probability approach:Again 1- the probability of the opposite event --> \(P=1-\frac{4}{5}*\frac{3}{4}=\frac{2}{5}\).
Answer: C.
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