Probability to shoot a target Three shooters aim at a target. The probability that they hit the target are $0.4$, $0.5$ and $0.7$,
respectively. Find the probability that the target is hit exactly once.
I don't know if this is just a conditional probability so we can something like:
$$P(A\mid B\cap C)+P(B\mid A\cap C)+P(C\mid A\cap B)$$ or it would be a Poisson model.
 A: Let $ A_i $ be the event that shooter number $ i $ hits the target. What youre looking for is just $ (A_1 \cap A^c_2  \cap A^c_3)\cup(A^c_1 \cap A_2  \cap A^c_3) \cup (A^c_1 \cap A^c_2  \cap A_3) $ Those are all disjoint events so it should'nt be hard to calculate.
A: Since there aren't too many shooters, it's easy to just list all the possibilities:
$$
\begin{split}
P(\text{target hit exactly once}) = &\\ 
&P(\text{A hits}) P(\text{B misses}) P(\text{C misses}) + \\
&P(\text{A misses}) P(\text{B hits}) P(\text{C misses}) +\\
&P(\text{A misses}) P(\text{B misses}) P(\text{C hits})\\
\end{split}
$$
A: Here is a neat way to organize the computation which not only yields the probability of exactly one hit but also the probability of exactly $n$ hits for $n=0,1,2,3$.
Let
$$\begin{align}
f_1(x) &= 0.6+0.4x \\
f_2(x) &= 0.5 + 0.5x \\
f_3(x) &= 0.3 + 0.7x
\end{align}$$
Now expand the product of these three polynomials:
$$f_1(x) f_2(x) f_3(x) = 0.09\, +0.36 x+0.41 x^2+0.14 x^3$$
The coefficient of $x^n$ in the result is the probability of exactly $n$ hits.  So the probability of exactly one hit is $0.36$.
