= Solution
Condition on whether the first selection was a red apple. Its <probability> is $3/18$. If it was, two red apples remain among seventeen fruits; otherwise three remain. The <law of total probability> gives
$$
\boxed{P(\text{second is red apple})
=\frac3{18}\frac2{17}+\frac{15}{18}\frac3{17}=\frac16}.
$$
The same result follows from <exchangeability>: before either selection is observed, each ordered draw position has the original red-apple fraction $3/18$, despite dependence between the draws.
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