Example 6.16. The probability of hitting the target with one shot is 0.8. Find the probability that in a series of four shots there will be: a) at least one hit; b) no less than three hits; c) no more than one hit.
Solution. Here $n=4, p=0.8, q=0.2$. a) Let's find the probability of the opposite event - in a series of four shots, there is not a single hit on the target:

$$
P_{4}(0)=C_{4}^{0} p^{0} q^{4}=0.2^{4}=0.0016
$$

From this, we find the probability of at least one hit on the target:

$$
P_{4}(k \geq 1)=1-0.0016=0.9984.
$$

b) The event $B$, which consists of at least three hits on the target in a series of four shots, means that there were either three hits (event $C$) or four (event $D$), i.e., $B=C+D$. Therefore, $P(B)=P(C)+P(D)$; hence,

$$
\begin{aligned}
P_{4}(k \geq 3) & =P_{4}(3)+P_{4}(4)=C_{4}^{3} p^{3} q^{1}+C_{4}^{4} p^{4} q^{0}= \\
& =4 \cdot 0.8^{3} \cdot 0.2+0.8^{4}=0.8192
\end{aligned}
$$

c) Similarly, the probability of hitting the target no more than once is calculated as follows:

$$
\begin{gathered}
P_{4}(k \leq 1)=P_{4}(0)+P_{4}(1)=0.0016+C_{4}^{1} p^{1} q^{3}= \\
=0.0016+4 \cdot 0.8 \cdot 0.2^{3}=0.2576
\end{gathered}
$$