Example 6. A worker services 3 machines on which identical parts are processed. The probability of defect for the first machine is 0.02, for the second - 0.03, for the third - 0.04. Processed parts are stored in one box. The productivity of the first machine is three times greater than that of the second, and the third is two times less than the second. What is the probability that a randomly taken part will be defective?
Solution. Let's introduce the following notations: event $A$ - "a defective part is taken"; hypothesis $H_{1}$ - "the part is made on the first machine", $H_{2}$ - "the part is made on the second machine", $H_{3}$ - "the part is made on the third machine".

Let $x$ be the productivity of the second machine, then $3 x$ is the productivity of the first machine, and $x / 2$ is the productivity of the third machine.

According to the problem statement, we have:

$$
\begin{aligned}
& P\left(H_{1}\right)=\frac{3 x}{3 x+x+x / 2}=\frac{3 x}{(9 / 2) x}=\frac{6}{9}=\frac{2}{3} \\
& P\left(H_{2}\right)=\frac{x}{(9 / 2) x}=\frac{2}{9}, \quad P\left(H_{3}\right)=\frac{x / 2}{(9 / 2) x}=\frac{1}{9} \\
& P\left(A / H_{1}\right)=0.02, \quad P\left(A / H_{2}\right)=0.03, \quad P\left(A / H_{3}\right)=0.04
\end{aligned}
$$

Using formula (1.9.1), we find the required probability

$$
P(A)=\frac{2}{3} \cdot 0.02+\frac{2}{9} \cdot 0.03+\frac{1}{9} \cdot 0.04=\frac{22}{900} \approx 0.024
$$