A football team's coach believes that his players kick the penalty with 95% certainty. What is the probability that out of five players, exactly three will miss?
Solution. Since the performance of the five players is independent of each other, we can apply the formula for the probability of the simultaneous occurrence of independent events $P\left(A_{1} ; A_{2} ; \ldots ; A_{n}\right)=P\left(A_{1}\right) \cdot P\left(A_{2}\right) \cdot \ldots \cdot P\left(A_{n}\right)$.

Let the five players be denoted by $A, B, C, D, E$. According to the above formula, the probability that exactly $C, D$, and $E$ miss their penalty kicks is $0.95 \cdot 0.95 \cdot 0.05 \cdot 0.05 \cdot 0.05$ (since if the probability that someone scores is 0.95, then the probability of the negation of this event, or that they miss the penalty, is $0.05$.

Since we can choose 3 people from 5 in more than one way, we need to multiply the obtained probability by $\binom{5}{3}$: $\binom{5}{3} \cdot 0.95^{2} \cdot 0.05^{3} \approx 0.00113$

Therefore, the probability that three out of five will miss their penalty kicks is approximately $0.00113$.