Example 6.11. There are three urns with balls. The first urn contains 4 white and 5 black balls, the second - 5 white and 4 black balls, the third - 6 white balls. Someone randomly selects one of the urns and draws a ball from it. Find the probability that: a) the ball will be white; b) the white ball is drawn from the second urn.
## Solution:

a) Let $A$ be the event that a white ball is drawn. Consider three hypotheses:

$$
\begin{aligned}
& H_{1} \text { - the first urn is chosen; } \\
& \mathrm{H}_{2} \text { - the second urn, } \text {; } \\
& H_{3} \text { - the third urn, . . }
\end{aligned}
$$

Since the urn from which the ball is drawn is chosen at random, we have

$$
P\left(H_{1}\right)=P\left(H_{2}\right)=P\left(H_{3}\right)=\frac{1}{3} .
$$

The conditional probabilities of event $A$ are respectively: $P\left(A / H_{1}\right)=\frac{4}{9}$ (the probability of drawing a white ball from the first urn), $P\left(A / H_{2}\right)=\frac{5}{9}$ (the probability of drawing a white ball from the second urn), $P\left(A / H_{3}\right)=1 \quad$ (the probability of drawing a white ball from the third urn).

From this, using the formula for total probability, we get:

$$
P(A)=\frac{1}{3} \cdot \frac{4}{9}+\frac{1}{3} \cdot \frac{5}{9}+\frac{1}{3} \cdot 1=\frac{1}{3} \cdot 2=\frac{2}{3}
$$

b) To determine the probability that the white ball was drawn from the second urn, we use Bayes' formula:

$$
P\left(H_{2} / A\right)=\frac{P\left(H_{2}\right) P\left(A / H_{2}\right)}{P(A)}=\frac{\frac{1}{3} \cdot \frac{5}{9}}{\frac{2}{3}}=\frac{5}{18}
$$