## Task 19/67

Two children are playing with a five-penny coin and a die. They agree: A throws the coin once, B throws the die three times. The one who rolls a "five" wins.

How are the winning chances distributed?
It is obvious that for $A$ the probability $P(5)=0.5$ is, since there are two possible cases, one of which is favorable. $B$ must have at least one five among the three rolls to win. Since the three events "5 on the $i$-th roll" ($i=1; 2; 3$) do not exclude each other, the addition rule can be applied. Therefore, one proceeds to the complementary event: $B$ loses if the three events "not 5 on the $i$-th roll" occur; for this (since they are independent of each other) the multiplication rule applies. Accordingly, the probability for $B$ to roll a five is

$$
P(5)=1-P(\overline{5})^{3}=1-\left(\frac{5}{6}\right)^3 \approx 0.42
$$

Thus, $B$ has significantly lower winning chances.