Solution
= Solution
Let $C_j$ identify the selected coin and let $H$ be the event of two heads. The prior coin <probabilities> are equal, and the <conditional probabilities> of $H$ are $1/9,1/4,4/9$. By <Bayes' theorem>,
$$
\boxed{\mathbb P(C_1\mid H)=
\frac{(1/3)(1/9)}{(1/3)(1/9+1/4+4/9)}
=\frac4{29}.}
$$
The full <posterior distribution> over the three biases is $(4,9,16)/29$.