Example 3. A batch of electric bulbs is manufactured by the first factory at $20 \%$, by the second factory at $30 \%$, and by the third factory at $50 \%$. The probabilities of producing defective bulbs are respectively: $q_{1}=0.01, q_{2}=0.005, q_{3}=0.006$. Find the probability that a randomly selected bulb from the batch will be standard.
Solution. We introduce the following notations: $A$ - "a standard bulb is taken from the batch", $H_{1}$ - "the taken bulb is manufactured by the first factory", $H_{2}$ - "by the second factory", $H_{3}$ - "by the third factory". We find the conditional probabilities $P\left(A / H_{i}\right)(i=1,2,3)$ using the formula

$$
P\left(A / H_{i}\right)=1-P\left(\bar{A} / H_{i}\right)
$$

where $\bar{A}$ is the event opposite to event $A$ (a non-standard bulb is taken):

$$
\begin{aligned}
& P\left(A / H_{1}\right)=1-P\left(\bar{A} / H_{1}\right)=1-0.01=0.99 \\
& P\left(A / H_{2}\right)=1-P\left(\bar{A} / H_{2}\right)=1-0.005=0.995 \\
& P\left(A / H_{3}\right)=1-P\left(\bar{A} / H_{3}\right)=1-0.006=0.994
\end{aligned}
$$

From the problem statement, it follows that

$$
P\left(H_{1}\right)=0.2, \quad P\left(H_{2}\right)=0.3, \quad P\left(H_{3}\right)=0.5
$$

In accordance with formula (1.9.1), we obtain

$$
\begin{aligned}
& P(A)=P\left(H_{1}\right) P\left(A^{\prime} / H_{1}\right)+P\left(H_{2}\right) P\left(A / H_{2}\right)+P\left(H_{3}\right) P\left(A / H_{3}\right)= \\
& =0.2 \cdot 0.99+0.3 \cdot 0.995+0.5 \cdot 0.994=0.198+0.2985+0.4970=0.9935
\end{aligned}
$$