Example 9. A coin is tossed 10 times. Find the probability that heads will appear: a) from 4 to 6 times; b) at least once.
Solution. According to formula (3.1.6) for $n=10, k_{1}=4, k_{2}=6, p=q=0.5$ we find

$$
P_{10}(4 \leq k \leq 6)=P_{10}(4)+P_{10}(5)+P_{10}(6)=\frac{210}{1024}+\frac{252}{1024}+\frac{210}{1024}=\frac{21}{32}
$$

According to formula (3.1.7) we obtain

$$
P_{10}(1 \leq k \leq 10)=1-\left(\frac{1}{2}\right)^{10}=\frac{1023}{1024}
$$