65. Suppose you have forgotten one digit of a phone number you need and dial it at random. What is the probability that you will have to make no more than two calls?
65. The considered experiment consists in randomly dialing a forgotten digit; if a wrong number is dialed, another digit is randomly dialed. Thus, there are two different types of outcomes here: either two numbers are dialed, or only one (the latter will be the case if the correct number is dialed the first time). We have no basis to consider any outcome of the first type and the outcome of the second type equally probable. To obtain a system of equally probable outcomes, it is convenient to assume conditionally that in all cases two numbers are dialed, i.e., that even if the correct number is dialed the first time, another number is randomly dialed afterward and it is asked whether the correct number was dialed (of course, in practice, there is no need to actually make the second call).

The total number of possible equally probable outcomes of a sequence of two telephone calls is $10 \cdot 9 = 90$ (the first time any of the 10 digits can be dialed, and the second time any of the remaining 9 digits). Favorable outcomes here, obviously, are nine outcomes where the correct digit is dialed the first time and any of the nine remaining digits the second time, and nine outcomes where any of the nine incorrect digits is dialed the first time and the correct digit the second time. Thus, the total number of favorable outcomes is $9 + 9 = 18$. Therefore, the desired probability is $\frac{18}{90} = \frac{1}{5} = 0.2$.