Example 2. Five fair coins are tossed. Find the probability that: exactly 2 heads appeared; more than one head appeared.
Solution. Let $X$ be the number of heads that appear in these tosses. In this case, $p=1 / 2$ and $q=1 / 2$.

$$
P(X=2)=P_{5}(2)=C_{5}^{2} p^{2} q^{s-2}=10 \cdot\left(\frac{1}{2}\right)^{2} \cdot\left(\frac{1}{2}\right)^{3}=10 \cdot \frac{1}{4} \cdot \frac{1}{8}=\frac{5}{16}=0.3125
$$

The random variable $X$ can take the following values $x_{1}=0$, $x_{2}=1, x_{3}=2, x_{4}=3, x_{5}=4, x_{6}=5$. Since $P(x>1)=1-P(x \leq 1)=$ $=1-P_{5}(0)-P_{5}(1)$, then

$$
P(x>1)=1-C_{s}^{0} p^{0} q^{5-0}-C_{s}^{1} p q^{s-1}=1-\left(\frac{1}{2}\right)^{5}-5\left(\frac{1}{2}\right)^{5}=\frac{26}{32}=\frac{13}{16}=0.8125
$$