Three shooters shoot at the same target, each of them shoots just once. The first one hits the target with a probability of $70\%$, the second one with a probability of $80\%$, and the third one with a probability of $90\%$. What is the probability that the shooters will hit the target
a) at least once?
b) at least twice?
I don't understand well this problem. If each shooter shoots at the target once, how can each of them shoot at it twice?
Do you know how to solve this problem?
Best Answer
Your solution to the first problem is correct.
Let $A$ be the event that the first shooter hits the target; let $B$ be the event that the second shooter hits the target; let $C$ be the event that the third shooter hits the target. If at least two shooters hit the target, then either exactly two of them hit the target or all three do. Thus, we must calculate $$\Pr(A)\Pr(B)\Pr(C^C) + \Pr(A)\Pr(B^C)\Pr(C) + \Pr(A^C)\Pr(B)\Pr(C) + \Pr(A)\Pr(B)\Pr(C)$$ given $\Pr(A) = 0.70$, $\Pr(B) = 0.80$, $\Pr(C) = 0.90$. Can you proceed?