A fair die is rolled once. If it shows 1 or 2, two balls are drawn without replacement from urn U1 containing 2 white and 3 black balls; otherwise…

Use Bayes' theorem. The prior for $U_1$ is $2/6=1/3$ and for $U_2$ is $2/3$. The chance of two white balls from $U_1$ is $\binom22/\binom52=1/10$, and from $U_2$ it is $\binom32/\binom52=3/10$. The joint probabilities are $\tfrac13\cdot\tfrac1{10}=\tfrac1{30}$ and $\tfrac23\cdot\tfrac3{10}=\tfrac15=\tfrac6{30}$. Therefore $p=\dfrac{1/30}{1/30+6/30}=\dfrac17$, and $1/p=7$. A common slip is to forget that the urn is chosen with unequal prior probabilities, which would give $p=1/4$ and the wrong answer $4$. The sanity check is that the posterior must be smaller than the prior of one third, because the white-rich urn U2 is far more likely to give two white balls; the value one seventh satisfies this.