 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 givesEquivalently, 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.