105. There are three batches of parts, each containing 20 parts. The number of standard parts in the first, second, and third batches is $20, 15, 10$, respectively. A part is randomly selected from a randomly chosen batch and turns out to be standard. The part is returned to the batch, and a second part is randomly selected from the same batch, which also turns out to be standard. Find the probability that the parts were drawn from the third batch.
Solution. Let $A$ be the event that in each of the two trials (with replacement) a standard part was drawn.

We can make three assumptions (hypotheses): $B_{1}$ - parts were drawn from the first batch; $B_{2}$ - parts were drawn from the second batch; $B_{3}$ - parts were drawn from the third batch.

The parts were drawn from a randomly chosen batch, so the probabilities of the hypotheses are the same:

$$
P\left(B_{1}\right)=P\left(B_{2}\right)=P\left(B_{3}\right)=1 / 3
$$

Let's find the conditional probability $P_{B_{1}}(A)$, i.e., the probability that two standard parts will be drawn sequentially from the first batch. This event is certain, as all parts in the first batch are standard, so

$$
P_{B_{1}}(A)=1
$$

Let's find the conditional probability $P_{B_{2}}(A)$, i.e., the probability that two standard parts will be drawn sequentially (with replacement) from the second batch:

$$
P_{B_{2}}(A)=15 / 20 \cdot 15 / 20=9 / 16
$$

Let's find the conditional probability $P_{B_{3}}(A)$, i.e., the probability that two standard parts will be drawn sequentially (with replacement) from the third batch:

$$
P_{B_{3}}(A)=10 / 20 \cdot 10 / 20=1 / 4
$$

The desired probability that both drawn standard parts were taken from the third batch, according to Bayes' formula, is

$$
\begin{gathered}
P_{A}\left(B_{3}\right)=\frac{P\left(B_{3}\right) \cdot P_{B_{3}}(A)}{P\left(B_{1}\right) \cdot P_{B_{1}}(A)+P\left(B_{2}\right) \cdot P_{B_{2}}(A)+P\left(B_{3}\right) \cdot P_{B_{3}}(A)}= \\
=\frac{1 / 3 \cdot 1 / 4}{1 / 3 \cdot 1+1 / 3 \cdot 9 / 16+1 / 3 \cdot 1 / 4}=4 / 29 .
\end{gathered}
$$