52. There are 10 identical urns, 9 of which contain 2 black and 2 white balls, and one contains 5 white and 1 black ball. A ball is drawn from a randomly selected urn. The drawn ball turned out to be white. What is the probability that this ball was drawn from the urn containing 5 white balls?
S o l u t i o n. There are 2 groups of urns with different compositions of balls; 9 of the existing urns belong to the first group, one urn - to the second group. The experiment consists of drawing a ball from a randomly selected urn. Let's consider the hypotheses:

$B_{1}$ - an urn from the first group is selected;

$B_{2}$ - an urn from the second group is selected;

and the event $A$ - drawing a white ball.

The probabilities of the hypotheses $B_{1}$ and $B_{2}$ are:

$P\left(B_{1}\right)=9 / 10=0.9, P\left(B_{2}\right)=1 / 10=0.1$. Hypotheses $B_{1}$ and $B_{2}$ form a complete group of mutually exclusive events, the sum of the probabilities of the hypotheses is $P\left(B_{1}\right)+P\left(B_{2}\right)=0.9+0.1=1$.

Event $A$ can occur only either together with event $B_{1}$ or with event $B_{2}$. Before the experiment, the conditional probability of event $A$, calculated on the condition that it occurred together with event $B_{1}$, is $P_{B_{1}}(A)=2 / 4=1 / 2$ (since in each urn of the first group, among the four balls, there are 2 white balls). Before the experiment, the conditional probability of event $A$, calculated on the condition that it occurred together with event $B_{2}$, is $P_{B_{2}}(A)=5 / 6$ (since in the urn of the second group, among the six balls, there are 5 white ones).

Suppose event $A$ occurred: a white ball was drawn from some urn. Let's reassess the probability of the second hypothesis. Using formula (17), we get

$$
P_{A}\left(B_{2}\right)=\frac{P\left(B_{2}\right) P_{B_{2}}(A)}{P\left(B_{1}\right) P_{B_{1}}(A)+P\left(B_{2}\right) P_{B_{2}}(A)}=\frac{0.1 \cdot \frac{5}{6}}{0.1 \cdot \frac{5}{6}+0.9 \cdot \frac{1}{2}}=0.15625
$$