# Three shots, two hits

### The problem

Three shooters fire to target. Probability of hit target for shooters 0.9, 0.8, 0.7 for first, second and third shooters accordingly.

What of probability of two hits to target?

### Solution

Express sought event `E`

though existing events `A`

, `B`

and `C`

:

$$\\ A - \texttt{is an event, 1-st shooter hit the target} \\ B - \texttt{is an event, 2-nd shooter hit the target} \\ C - \texttt{is an event, 3-rd shooter hit the target} \\ E - \texttt{is an event, two shooters hit the target} \\ P(A) = 0.9 \\ P(B) = 0.8 \\ P(C) = 0.7 \\ P(E) = (A\cap B\cap \bar{C})\cup(A\cap \bar{B}\cap C)\cup(\bar{A}\cap B\cap C) \\ P(E) = 0.9\times 0.8\times(1-0.7)+0.9\times(1-0.8)\times0.7+(1-0.9)\times0.8\times0.7 = 0.398$$