Solution
= Solution
Condition on an apple occupying position two in a uniformly random ordering. The nine other positions then contain four apples and five bananas, each equally likely to occupy position three. The <conditional probability> is therefore
$$
\boxed{P(A_3\mid A_2)=\frac49.}
$$
This uses <exchangeability> of the unconditioned positions; there is no need to assume the first lunch contained a particular fruit.