Probability to shoot a target

Question

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.

Answer 1

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.

Answer 2

Since there aren't too many shooters, it's easy to just list all the possibilities: $$ \begin{split} P(\text{target hit exactly once}) = &\\ &P(\text{A hits}) P(\text{B misses}) P(\text{C misses}) + \\ &P(\text{A misses}) P(\text{B hits}) P(\text{C misses}) +\\ &P(\text{A misses}) P(\text{B misses}) P(\text{C hits})\\ \end{split} $$

Answer 3

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 $0.36$.