Example 5. Parts from three machines arrive at the assembly. It is known that the first machine produces $0.1\%$ of defective parts, the second - $0.2\%$, the third - $0.3\%$. Find the probability of a defective part arriving at the assembly, if the first machine supplied 1000, the second - 2000, and the third - 3000 parts.
Solution. We introduce the following notations: event $A$ - "a defective part has arrived"; hypothesis $H_{1}$ - "the part was manufactured on the first machine", $H_{2}$ - "on the second", $H_{3}$ - "on the third".

From the problem statement, it follows that

$$
\begin{aligned}
& P\left(H_{1}\right)=\frac{1000}{6000}=\frac{1}{6}, \quad P\left(H_{2}\right)=\frac{2000}{6000}=\frac{1}{3}, \quad P\left(H_{3}\right)=\frac{3000}{6000}=\frac{1}{2} \\
& P\left(A / H_{1}\right)=0.001, \quad P\left(A / H_{2}\right)=0.002, \quad P\left(A / H_{3}\right)=0.003 .
\end{aligned}
$$

In accordance with formula (1.9.1), we find the required probability

$$
\begin{aligned}
& P(A)=P\left(H_{1}\right) P\left(A / H_{1}\right)+P\left(H_{2}\right) P\left(A / H_{2}\right)+P\left(H_{3}\right) P\left(A / H_{3}\right)= \\
& =\frac{1}{6} \cdot 0.001+\frac{1}{3} \cdot 0.002+\frac{1}{2} \cdot 0.003=\frac{14}{6000} \approx 0.0023
\end{aligned}
$$