43. The subscriber has forgotten the last digit of the phone number and therefore dials it at random. What is the probability that they will have to dial the number no more than three times?
S o l u t i o n. Let us denote the event:

C - the subscriber will have to dial the number no more than three times.

This event consists in the subscriber having to dial the number either once, or twice, or three times. Let us consider the following events:

$C_{1}$ - the subscriber will dial the number once;

$C_{2}$ - the subscriber will dial the number twice;

$C_{3}$ - the subscriber will dial the number three times;

$\overline{C_{1}}$ - the correct digit is not dialed the first time;

$A$ - the correct digit is dialed the second time;

$\bar{A}$ - the correct digit is not dialed the second time;

$B$ - the correct digit is dialed the third time.

The event $C$ is the sum of the mutually exclusive events $C_{1}, C_{2}$, and $C_{3}: C=C_{1}+C_{2}+C_{3}$. The probability of the event $C_{1}$, according to formula (1), is $P\left(C_{1}\right)=1 / 10$.

The event $C_{2}$ consists in the correct digit not being dialed the first time, but being dialed the second time. This means that $C_{2}$ is the product of the events $\overline{C_{1}}$ and $A: C_{2}=\overline{C_{1}} \cdot A$. The probability of the event $\overline{C_{1}}$ is $P\left(\overline{C_{1}}\right)=9 / 10$. The event $A$ is dependent on the event $\overline{C_{1}}$; the conditional probability $P_{C_{1}}(A)=1 / 9$. The probability of the event $C_{2}$ will be found using the theorem of multiplication of probabilities of dependent events. Applying formula (12), we get: $P\left(C_{2}\right)=P\left(\bar{C}_{1}\right) \cdot P_{\bar{C}_{1}}(A)=\frac{9}{10} \cdot \frac{1}{9}=\frac{1}{10}$.

The event $C_{3}$ consists in the correct digit not being dialed the first and second times, but being dialed the third time. This means that $C_{3}$ is the product of the dependent events $\overline{C_{1}}, \bar{A}$, and $B: C_{3}=$ $=\overline{C_{1}} \cdot \overline{\boldsymbol{A}} \cdot \boldsymbol{B}$. The conditional probability $P_{\bar{C}_{1}}(\bar{A})=8 / 9$; the conditional probability $P_{C_{1} \cdot A}(B)=1 / 8$. The probability of the event $C_{3}$ will be found using the theorem of multiplication of probabilities of dependent events. Applying formula (14), we get:

$$
P\left(C_{3}\right)=P\left(\overline{C_{1}}\right) \cdot P_{\overline{C_{1}}}(\bar{A}) \cdot P_{\bar{C}_{1 \overline{ }}}(B)=\frac{9}{10} \cdot \frac{8}{9} \cdot \frac{1}{8}=\frac{1}{10}
$$

The desired probability of the event $C$ will be found using the theorem of addition of probabilities of mutually exclusive events. According to formula (9), this probability is equal to

$$
P(C)=P\left(C_{1}\right)+P\left(C_{2}\right)+P\left(C_{3}\right)=0,1+0,1+0,1=0,3
$$