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.