49. The probabilities of occurrence of each of two independent events $A_{1}$ and $A_{2}$ are $p_{1}$ and $p_{2}$, respectively. Find the probability of the occurrence of only one of these events.
Solution. Let's introduce the notation for events: $B_{1}$ - only event $A_{1}$ occurs; $B_{2}$ - only event $A_{2}$ occurs.

The occurrence of event $B_{1}$ is equivalent to the occurrence of event $A_{1} \bar{A}_{2}$ (the first event occurs and the second does not), i.e., $B_{1}=A_{1} \bar{A}_{2}$. The occurrence of event $B_{2}$ is equivalent to the occurrence of event $\bar{A}_{1} A_{2}$ (the second event occurs and the first does not), i.e., $B_{2}=\bar{A}_{1} A_{2}$:

Thus, to find the probability of only one of the events $A_{1}$ and $A_{2}$ occurring, it is sufficient to find the probability of one, regardless of which, of the events $B_{1}$ and $B_{2}$. Events $B_{i}$ and $B_{3}$
are mutually exclusive, so the addition theorem can be applied:

$$
P\left(B_{1}+B_{2}\right)=P\left(B_{1}\right)+P\left(B_{2}\right)
$$

It remains to find the probabilities of each of the events $B_{\mathbf{i}}$ and $B_{\mathbf{g}}$. Events $A_{1}$ and $A_{2}$ are independent, hence events $A_{1}$ and $\bar{A}_{2}$, as well as $\bar{A}_{1}$ and $A_{2}$, are independent, so the multiplication theorem can be applied:

$$
\begin{aligned}
& P\left(B_{1}\right)=P\left(A_{1} \bar{A}_{2}\right)=P\left(A_{1}\right) \cdot A\left(\bar{A}_{2}\right)=p_{1} q_{2} ; \\
& P\left(B_{2}\right)=P\left(\bar{A}_{1} A_{2}\right)=P\left(\bar{A}_{1}\right) \cdot P\left(A_{2}\right)=q_{1} p_{2}
\end{aligned}
$$

Substituting these probabilities into the relation (*), we find the desired probability of only one of the events $A_{1}$ and $A_{2}$ occurring:

$$
P\left(B_{1}+B_{2}\right)=p_{1} q_{2}+q_{1} p_{2}
$$