= Solution
If the selected coin has bias $p$, the other two biases sum to $3/2-p$. A uniform choice between those two coins therefore has conditional head <probability> $3/4-p/2$. Average over the <posterior distribution>, using the previous <posterior predictive probability>:
$$
\boxed{\mathbb P(\text{head on another coin}\mid H)
=\frac34-\frac12\frac{33}{58}
=\frac{27}{58}.}
$$
The observation makes a high-bias coin more likely to have been removed from the remaining pair.
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