Example 15. Two equally strong chess players are playing chess. What is more likely: a) to win one game out of two or two games out of four? b) to win at least two games out of four or at least three games out of five? Draws are not considered.
Solution. Since the chess players are of equal strength, the probability of winning $p=1 / 2$; therefore, the probability of losing $q$ is also
equal to $1 / 2$. In all games, the probability of winning is constant and it does not matter in what sequence the games will be won, so the Bernoulli formula can be applied.

Using formulas (3.1.1) and (3.1.10), we find the specified probabilities:

$$
\begin{aligned}
& P_{2}(1)=C_{2}^{1} p q=2 \cdot \frac{1}{2} \cdot \frac{1}{2}=\frac{1}{2} \\
& P_{4}(2)=C_{4}^{2} p^{2} q^{2}=6 \cdot \frac{1}{2^{2}} \cdot \frac{1}{2^{2}}=\frac{3}{8} \\
& P_{4}(3)=C_{4}^{3} p^{3} q=4 \cdot \frac{1}{2^{3}} \cdot \frac{1}{2}=\frac{1}{4} \\
& P_{4}(4)=C_{4}^{4} p^{4} q^{0}=1 \cdot \frac{1}{2^{4}}=\frac{1}{16} \\
& P_{4}(2)+P_{4}(3)+P_{4}(4)=\frac{3}{8}+\frac{1}{4}+\frac{1}{16}=\frac{11}{16} \\
& P_{5}(3)=C_{5}^{3} p^{3} q^{2}=10 \cdot \frac{1}{2^{3}} \cdot \frac{1}{2^{2}}=\frac{5}{16} \\
& P_{5}(4)=C_{5}^{4} p^{4} q=5 \cdot \frac{1}{2^{4}} \cdot \frac{1}{2}=\frac{5}{32} \\
& P_{5}(5)=C_{5}^{5} p^{5} q^{0}=1 \cdot \frac{1}{2^{5}} \cdot 1=\frac{1}{32} \\
& P_{5}(3)+P_{5}(4)+P_{5}(5)=\frac{5}{16}+\frac{5}{32}+\frac{1}{32}=\frac{16}{32}=\frac{8}{16}
\end{aligned}
$$

Since $P_{2}(1)=\frac{1}{2}, P_{4}(2)=\frac{3}{8}$ and $P_{2}(1)>P_{4}(2)$, it is more likely to win one game out of two.

Since $\quad P_{4}(2)+P_{4}(3)+P_{4}(4)=\frac{11}{16}, \quad P_{5}(3)+P_{5}(4)+P_{5}(5)=\frac{8}{16} \quad$ and $\frac{11}{16}>\frac{8}{16}$, it is more likely to win at least two games out of four.

Remark. The probability of winning at least two games out of four can also be calculated in another way: $P_{4}(2)+P_{4}(3)+P_{4}(4)=1-\left[P_{4}(0)+P_{4}(1)\right]$. 180

Since $P_{4}(0)=\frac{1}{16}, P_{4}(1)=\frac{1}{4}$, then

$$
P_{4}(2)+P_{4}(3)+P_{4}(4)=1-\left(\frac{1}{16}+\frac{1}{4}\right)=\frac{11}{16}
$$