1565. There are 12 urns, of which 6 urns (composition $A_{1}$) contain 3 white and 4 black balls, 3 urns (composition $A_{2}$) contain 2 white and 8 black balls, 2 urns (composition $A_{3}$) contain 6 white and 1 black ball, and 1 urn (composition $A_{4}$) contains 4 white and 3 black balls. A ball is drawn from a randomly selected urn. What is the probability that the ball was drawn from an urn of composition $A_{3}$, given that it is white?
Solution. Let's introduce notation. Let event $B$ be the event that the ball drawn is white. According to the condition, $P\left(A_{1}\right)=\frac{6}{12}=\frac{1}{2}$, $P\left(A_{2}\right)=\frac{3}{12}=\frac{1}{4}, \quad P\left(A_{3}\right)=\frac{2}{12}=\frac{1}{6}, \quad P\left(A_{4}\right)=\frac{1}{12} ; \quad P\left(B / A_{1}\right)=$ $=\frac{3}{3+4}=\frac{3}{7}, \quad P\left(B / A_{2}\right)=\frac{2}{2+8}=\frac{1}{5}, \quad P\left(B / A_{3}\right)=\frac{6}{6+1}=\frac{6}{7}$, $P\left(B / A_{4}\right)=\frac{4}{4+3}=\frac{4}{7}$. By Bayes' formula, we find the required probability

$$
\begin{aligned}
& P\left(A_{3} / B\right)=\frac{P\left(A_{3}\right) P\left(B / A_{3}\right)}{P\left(A_{1}\right) P\left(B / A_{1}\right)+P\left(A_{2}\right) P\left(B / A_{2}\right)+P\left(A_{3}\right) P\left(B / A_{3}\right)+P\left(A_{4}\right) P\left(B / A_{4}\right)}= \\
& =\frac{\frac{1}{6} \cdot \frac{6}{7}}{\frac{1}{2} \cdot \frac{3}{7}+\frac{1}{4} \cdot \frac{1}{5}+\frac{1}{6} \cdot \frac{6}{7}+\frac{1}{12} \cdot \frac{4}{7}}=\frac{\frac{1}{7}}{\frac{3}{14}+\frac{1}{20}+\frac{1}{7}+\frac{1}{21}}=0.314 .
\end{aligned}
$$