53. The first machine supplies 80% of the parts for assembly, while the second machine supplies 20% of the same parts. The defect rate on the first machine is 1%, and on the second machine, it is 5%. A checked part turned out to be defective. Is it more likely that this part was manufactured on the first machine or on the second machine?
Solution. The test consists of checking the quality of a part. Consider the events:

$B_{1}$ - the part checked was manufactured on the first machine;

$B_{2}$ - the part checked was manufactured on the second machine;

$A$ - the part checked is defective.

The unconditional probabilities of the hypotheses - events $B_{1}$ and $B_{2}$ before the test are: $P\left(B_{1}\right)=0.8, P\left(B_{2}\right)=0.2$. The hypotheses $B_{1}$ and $B_{2}$ form a complete group of mutually exclusive events, the sum of the probabilities of these events is $P\left(B_{1}\right)+P\left(B_{2}\right)=1$.

On the first machine, the defect rate is $1\%$, so $P_{B_{1}}(A)=0.01$. On the second machine, the defect rate is $5\%$, so $P_{B_{2}}(A)=0.05$.

Event $A$ can occur either together with event $B_{1}$ or with event $B_{2}$.

Suppose event $A$ has occurred. Let's reassess the probabilities of the hypotheses. Using Bayes' formula, we find:

$$
\begin{aligned}
& P_{A}\left(B_{1}\right)=\frac{P\left(B_{1}\right) \cdot P_{B_{1}}(A)}{P\left(B_{1}\right) \cdot P_{B_{1}}(A)+P\left(B_{2}\right) \cdot P_{B_{2}}(A)}=\frac{0.8 \cdot 0.01}{0.8 \cdot 0.01+0.2 \cdot 0.05}=\frac{4}{9} \\
& P_{A}\left(B_{2}\right)=\frac{P\left(B_{2}\right) \cdot P_{B_{2}}(A)}{P\left(B_{1}\right) \cdot P_{B_{1}}(A)+P\left(B_{2}\right) \cdot P_{B_{2}}(A)}=\frac{0.2 \cdot 0.05}{0.8 \cdot 0.01+0.2 \cdot 0.05}=\frac{5}{9} .
\end{aligned}
$$

It is obtained that $P_{A}\left(B_{2}\right)>P_{A}\left(B_{1}\right)$. Thus, it is more likely that the part checked, which turned out to be defective, was manufactured on the second machine.