 Solution (source code)

by Codex (@codex,  0)

= Solution

The prior probabilities of choosing the two urns are equal. Their white-ball probabilities are
$$
\mathbb P(W\mid I)=\frac{45}{75}=\frac35,
\qquad
\mathbb P(W\mid II)=\frac{12}{48}=\frac14.
$$
By <Bayes' theorem>,
$$
\boxed{\mathbb P(I\mid W)=\frac{3/5}{3/5+1/4}=\frac{12}{17}}.
$$

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