Example 6.10. A factory produces a certain type of product; each product has a defect with a probability of $p_{1}=0.1$. Each product is inspected by one inspector; he detects an existing defect with a probability of $p_{2}=0.8$, and if the defect is not detected, the product is released into the finished product. In addition, the inspector may incorrectly reject a product that does not have a defect; the probability of this is $p_{3}=0.3$. Find the probabilities of the following events:

$A_{1}$ - the product will be incorrectly rejected;

$A_{2}$ - the product will be released into the finished product with a defect;

$A_{3}$ - the product will be rejected.
Solution. Consider the following events:

$B_{1}$ - the item has a defect;

$B_{2}$ - the inspector will detect the existing defect;

$B_{3}$ - the inspector will reject an item that does not have a defect.

According to the problem, $\quad P\left(B_{1}\right)=p_{1}=0.1 ; \quad P\left(B_{2}\right)=p_{2}=0.8 ;$ $P_{\bar{B}_{1}}\left(B_{3}\right)=p_{3}=0.3$. The event $A_{1}$ by definition means: “the item does not have a defect and the item will be rejected by the inspector”, i.e., $A_{1}=\bar{B}_{1} \cdot B_{3}$. Then

$$
P\left(A_{1}\right)=P\left(\bar{B}_{1} \cdot B_{3}\right)=P\left(\bar{B}_{1}\right) \cdot P_{\bar{B}_{1}}\left(B_{3}\right)=\left(1-p_{1}\right) \cdot p_{3}=0.9 \cdot 0.3=0.27
$$

The event $A_{2}$ by definition means: “the item has a defect and the inspector does not detect the defect”, i.e., $A_{2}=B_{1} \cdot \bar{B}_{2}$. Then

$$
P\left(A_{2}\right)=P\left(B_{1}\right) \cdot P\left(\bar{B}_{2}\right)=p_{1} \cdot\left(1-p_{2}\right)=0.1 \cdot 0.2=0.02
$$

since events $B$ and $\bar{B}_{2}$ are independent.

The event $A_{3}$ by definition means: “the item has a defect and the inspector detects the defect or the item does not have a defect and the inspector rejects the item”, i.e.,

$$
A_{3}=B_{1} \cdot B_{2}+\bar{B}_{1} \cdot B_{3}
$$

and

$$
\begin{aligned}
P\left(A_{3}\right)= & P\left(B_{1} \cdot B_{2}+\bar{B}_{1} \cdot B_{3}\right)=P\left(B_{1}\right) \cdot P\left(B_{2}\right)+P\left(\bar{B}_{1}\right) \cdot P_{\bar{B}_{1}}\left(B_{3}\right)= \\
= & P_{1} \cdot P_{2}+\left(1-p_{1}\right) \cdot p_{3}=0.1 \cdot 0.8+0.9 \cdot 0.3=0.35 .
\end{aligned}
$$