110. Two equally skilled chess players are playing chess. What is more likely: to win two out of four games or three out of six (draws are not considered)?
Solution. The chess players are of equal strength, so the probability of winning $p=1 / 2$; consequently, the probability of losing $q$ is also $1 / 2$. Since the probability of winning is constant in all games and the order in which the games are won does not matter, the Bernoulli formula can be applied.

Let's find the probability that two games out of four will be won:

$$
P_{4}(2)=C_{4}^{2} p^{2} q^{2}=4 \cdot 3 /(1 \cdot 2) \cdot(1 / 2)^{2} \cdot(1 / 2)^{2}=6 / 16
$$

Let's find the probability that three games out of six will be won:

$$
P_{6}(3)=C_{6}^{3} p^{3} q^{3}=6 \cdot 5 \cdot 4 /(1 \cdot 2 \cdot 3) \cdot(1 / 2)^{3} \cdot(1 / 2)^{3}=5 / 16
$$

Since $P_{4}(2)>P_{6}(3)$, it is more likely to win two games out of four than three out of six.