1538. Three shooters shoot at a target independently of each other. The probability of hitting the target for the first shooter is 0.6, for the second - 0.7, and for the third - 0.75. Find the probability of at least one hit on the target if each shooter makes one shot.
Solution. Let $A, B, C$ be the events that the first, second, and third shooter hit the target, respectively. From the problem, we have $P(A)=0.6, P(B)=0.7$, 
$P(C)=0.75$. We need to find the probability of at least one hit, i.e., $P(A+B+C)$:

$$
\begin{aligned}
P(A+B+C) & =1-P(\bar{A}) P(\bar{B}) P(\bar{C})=1-(1-0.6)(1-0.7)(1- \\
& -0.75)=1-0.4 \cdot 0.3 \cdot 0.25=0.97
\end{aligned}
$$