Task 7. A subscriber has forgotten the last digit of the phone number and therefore dials it at random. Determine the probability that he will need to call no more than three times to get the correct connection.
Solution. Let's denote the events:

$A_{1}$ - the subscriber got through on the first attempt;

$A_{2}$ - the subscriber got through on the second attempt;

$A_{3}$ - the subscriber got through on the third attempt.

Then $A=A_{1}+\overline{A_{1}} A_{2}+\overline{A_{1}} \overline{A_{2}} A_{3}$ - the subscriber used no more than three attempts to dial the correct number. $A$ is the sum of three mutually exclusive events.

When guessing the last digit of the number, the subscriber has 10 digits from 0 to 9 at their disposal, one of which is correct. Therefore, $p\left(A_{1}\right)=\frac{1}{10}, p\left(\overline{A_{1}}\right)=\frac{9}{10}$. If the first digit dialed is incorrect, then for guessing, there are already 9 digits left, one of which is correct. Thus, $p\left(A_{2} / \bar{A}_{1}\right)=\frac{1}{9}, p\left(\bar{A}_{2} / \bar{A}_{1}\right)=\frac{8}{9}$.

If the second attempt also results in an incorrect digit, then 8 digits remain at the subscriber's disposal, one of which is correct, i.e., $p\left(A_{3} / \bar{A}_{1} \bar{A}_{2}\right)=\frac{1}{8}$.

By formulas (6) and (9) we have:

$$
\begin{aligned}
p(A)= & p\left(A_{1}\right)+p\left(\overline{A_{1}} A_{2}\right)+p\left(\overline{A_{1}} \overline{A_{2}} A_{3}\right)=p\left(A_{1}\right)+p\left(\overline{A_{1}}\right) \cdot p\left(A_{2} / \overline{A_{1}}\right)+ \\
& +p\left(\overline{A_{1}}\right) \cdot p\left(\overline{A_{2}} / \overline{A_{1}}\right) p\left(A_{3} / \overline{A_{1}} \overline{A_{2}}\right)=\frac{1}{10}+\frac{9}{10} \cdot \frac{1}{9}+\frac{9}{10} \cdot \frac{8}{9} \cdot \frac{1}{8}=0.3
\end{aligned}
$$

Answer. 0.3 .