Example 3. Products from three factories are delivered to a warehouse, where products from the first factory account for $20 \%$, the second - $46 \%$, and the third $34 \%$. It is known that the average percentage of non-standard items for the first factory is $3 \%$, for the second $-2 \%$, and for the third - $1 \%$. Find the probability that a randomly selected non-standard item was produced by the first factory.
Solution. Let $A$ be the event that a non-standard item is taken, and let $H_{1}, H_{2}, H_{3}$ be the hypotheses that the item taken was manufactured at the first, second, and third factory, respectively.

From the problem statement, it follows that

$$
\begin{aligned}
& P\left(H_{1}\right)=0.20, \quad P\left(H_{2}\right)=0.46, \quad P\left(H_{3}\right)=0.34 \\
& P\left(A / H_{1}\right)=0.03, \quad P\left(A / H_{2}\right)=0.02, \quad P\left(A / H_{3}\right)=0.01
\end{aligned}
$$

Since in this case

$$
P(A)=0.20 \cdot 0.03 + 0.46 \cdot 0.02 + 0.34 \cdot 0.01 = 0.0186
$$

then, according to formula (1.1,0.2), we find the required probability

$$
P\left(H_{1} / A\right)=\frac{0.20 \cdot 0.03}{P(A)}=\frac{0.006}{0.0186} \approx 0.322
$$

Remark. Similarly, the probabilities are found as follows:

$$
\begin{aligned}
& P\left(H_{2} / A\right)=\frac{0.46 \cdot 0.02}{P(A)}=\frac{0.0092}{0.0186} \approx 0.495, \\
& P\left(H_{3} / A\right)=\frac{0.34 \cdot 0.01}{P(A)}=\frac{0.0034}{0.0186} \approx 0.183
\end{aligned}
$$