A fair die is rolled once and shows the number n. Then n fair coins are tossed. Given that exactly two heads appear, the probability that n was a…

Let $H$ be the event of exactly two heads. $P(H\mid n)=\binom{n}{2}/2^{n}$, and each $n$ has prior $1/6$. Exactly two heads has probability $0$ for $n=1$, and for $n=2,\dots,6$ it is $\frac14,\ \frac38,\ \frac38,\ \frac{5}{16},\ \frac{15}{64}$ (from $\binom n2/2^n$). Multiplying by $\frac16$: the weights are $\frac1{24},\frac1{16},\frac1{16},\frac5{96},\frac5{128}$ for $n=2,\dots,6$. Their total is $P(H)=\frac{33}{128}$. The primes $2,3,5$ contribute $\frac1{24}+\frac1{16}+\frac5{96}=\frac{5}{32}$. By Bayes' theorem the required probability is $\frac{5/32}{33/128}=\frac{20}{33}$. The option $\frac{5}{32}$ is the joint probability, not the conditional one, and $\frac{5}{11}$ and $\frac{13}{33}$ come from mishandling the weights of $n=4$ and $n=6$. Exact fraction arithmetic gives $20/33$.