67. Each exam ticket includes two theoretical questions and one problem. In total, 28 tickets have been prepared. Calculate the probability that a student, randomly drawing a ticket, will answer all questions, if the student has prepared 50 theoretical questions and 22 problems:
Solution. The full answer to the ticket consists of the product of two events: the student will simultaneously answer two questions (event $A$) and solve a problem (event $B$). Let's calculate the probabilities of these events.

The number of all possible combinations of 2 questions out of 56 is

$$
C_{56}^{2}=\frac{56!}{54!\cdot 2!}=\frac{54!55 \cdot 56}{54!2 \cdot 1}=1540
$$

Since the student has prepared only 50 questions, the number of outcomes favorable to event $A$ is

$$
C_{50}^{2}=\frac{50!}{48!\cdot 2!}=\frac{48!\cdot 49 \cdot 50}{48!\cdot 1 \cdot 2}=1225
$$

Let's calculate the probability of event $A$:

$$
P(A)=\frac{C_{50}^{2}}{C_{58}^{2}}=\frac{1225}{1540}
$$

The probability of event $B$ is determined by the fact that the student knows 22 problems out of 28 possible:

$$
P(B)=\frac{22}{28}=\frac{11}{14}
$$

Since events $A$ and $B$ are independent and must occur simultaneously, we have

$$
P(A B)=P(A) P(B)=\frac{1225 \cdot 11}{1540 \cdot 14}=0.625
$$