89. A white ball is placed into an urn containing two balls, after which one ball is randomly drawn from it. Find the probability that the drawn ball will be white, if all possible assumptions about the initial composition of the balls (by color) are equally likely.
Solution. Let $A$ be the event of drawing a white ball. The following hypotheses about the original composition of the balls are possible: $B_{1}$ - no white balls, $B_{2}$ - one white ball, $B_{3}$ - two white balls.

Since there are three hypotheses, and by condition they are equally probable, and the sum of the probabilities of the hypotheses is one (since they form a complete group of events), the probability of each hypothesis is $1 / 3$, i.e., $P\left(B_{1}\right)=P\left(B_{2}\right)=P\left(B_{3}\right)=1 / 3$.

The conditional probability that a white ball will be drawn, given that there were no white balls initially in the urn, $P_{B_{1}}(A)=1 / 3$.

The conditional probability that a white ball will be drawn, given that there was one white ball initially in the urn, $P_{B_{2}}(A)=2 / 3$.

The conditional probability that a white ball will be drawn, given that there were two white balls initially in the urn, $P_{B_{3}}(A)=3 / 3=1$.

The desired probability that a white ball will be drawn is found using the formula for total probability:

$$
\begin{gathered}
P(A)=P\left(B_{1}\right) \cdot P_{B_{1}}(A)+P\left(B_{2}\right) \cdot P_{B_{3}}(A) \dot{+} \\
+P\left(B_{3}\right) \cdot P_{B_{3}}(A)=1 / 3 \cdot 1 / 3+1 / 3 \cdot 2 / 3+1 / 3 \cdot 1=2 / 3
\end{gathered}
$$