= Solution
Let $C$ denote a green pear on the first draw and $D$ a red apple on the second. The two categories are disjoint, so after $C$ there are still three red apples among seventeen remaining fruits. Thus $P(C\cap D)=(8/18)(3/17)$. Using $P(D)=1/6$ from (c), <conditional probability> gives
$$
\boxed{P(C\mid D)=\frac{(8/18)(3/17)}{1/6}=\frac8{17}}.
$$
Equivalently, conditioning on the second fruit being one particular category removes a red apple and leaves the other seventeen fruits symmetrically eligible for the first position.
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