Two riflemen A and B shoot at a target simultaneously. A has a 0.8 chance of hitting and B has a 0.9 chance What is the probability that the target will be hit? Assume independence.
$P( X\ge 1) $ is the probability that the target will be hit at least once The complement is $P(X = 0)$, probability that the target won't be hit. I calculated that to be $(1-0.8)\times (1-0.9) = 0.02$. So the solution is $1-0.02 = 0.98$? 
 A: Yes, that's a correct argument. (Assuming independence, which is not quite right in the real world because their bullets could collide in midair!)
A: You can calculate the probability that either hit by calculating the probability that both do not hit and subtracting that from 1. 
The probability of $A$ not hitting the target is $1 - 0.8 = 0.2$ and the probability of $B$ not hitting the target is $1 - 0.9 = 0.1$, so the probability they both miss is the product, $0.02$, so the probability that at least one of them hits the target is $0.98$
Another way to think of it is to deal with the cases independently. $A$ hits the target with $0.8$ probability. But what about the $0.2$ probability he doesn't? In that case, $B$ can hit the target with probability $0.9$ for an overall probability of $0.2 * 0.9 = .18$ 
So the probability that the target is hit is $0.8 + .18 = 0.98$
A: A little explanation (or easy-to-understand) of the result can be taken as:
$$
\begin{align}
P\left( X \ge 1 \right) &= P\left( X = 2 \right) + P\left( X = 1 \right) \\ 
&= (0.9 \times 0.8) + P(\text{ A hit the target, B didn't }) + P(\text{ B hit the target, A didn't }) \\
&= 0.72 + (0.90 \times 0.02) + (0.80 \times 0.01) \\
&= 0.72 + 0.18 + 0.08 \\
&= 0.98
\end{align}$$
