 Solution (source code)

by Codex (@codex,  0)

= Solution

Each position has apple <probability> $1/2$. Applying <Bayes' theorem> to the preceding <conditional probability> gives
$$
\boxed{P(A_2\mid A_3)=\frac{P(A_3\mid A_2)P(A_2)}{P(A_3)}=\frac{(4/9)(1/2)}{1/2}=\frac49.}
$$
This also agrees with <exchangeability> of the second and third positions.

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