Three shooters $A,B,C$ aim at a target. Each of them shoots one. The accuracy of the $A$ shooter is equal to $0.6$, $B$ is $0.7$, $C$ is $0.8$. What's the probability that $A$ hit the target if:
- the target was hit once,
- the target was hit twice,
- the target was hit at least once,
- the target was hit at least twice.
It's obvious that $$ P(A)=0.6$$ $$P(B)=0.7$$ $$ P(C)=0.6$$ where $P(\cdot)$ is the probability that the shooter hit the target.
For 1) I have: $$P(A|T_{1})=\frac{P(T_{1}|A)\cdot P(A)}{P(T_{1}|A)\cdot P(A) + P(T_{1}|B)\cdot P(B) + P(T_{1}|C)\cdot P(C)} $$ where $P(T_{1}|A)=0.6*(1-0.7)*(1-0.8)$ and respectively for the other ones.
For 2) $$P(A|T_{2})=\frac{P(T_{2}|A\cap B)\cdot P(A \cap B) + P(T_{2}|A\cap C)\cdot P(A \cap C)}{...}$$ Now, I'm kind of lost. What I wanted to obtain is the probability that A hit the target and the target was hit twice - so either B or C can aim with A. But what's $P(A \cap B)$? Maybe I want $P(A \cup B)$?
For 3) My first thought was to make this with the opposite event. So the target wasn't hit at all. But I can't imagine how to appy Bayes theorem to this case. I mean if the target wasn't hit then the probability that A hit the target is 0. Something doesn't feel right. For 4) I have a similar problem.
First question:
$$\mathbb{P}[A|\text{1 hit}]=\frac{\mathbb{P}[A]\mathbb{P}[\overline{B}]\mathbb{P}[\overline{C}]}{\mathbb{P}[A]\mathbb{P}[\overline{B}]\mathbb{P}[\overline{C}]+\mathbb{P}[\overline{A}]\mathbb{P}[B]\mathbb{P}[\overline{C}]+\mathbb{P}[\overline{A}]\mathbb{P}[\overline{B}]\mathbb{P}[C]}=\frac{36}{36+56+96}=\frac{9}{47}$$
Second question:
Consider that if the target was hit twice the following cases are possible
$$AB,AC,BC$$
Thus the solution is very similar to 1)
$$\mathbb{P}[A|\text{2 hits}]=\frac{0.6\times0.7\times0.2+0.6\times0.3\times0.8}{0.6\times0.7\times0.2+0.6\times0.3\times0.8+0.4\times0.7\times0.8}=\frac{228}{452}=\frac{57}{113}$$
Third question:
$$\mathbb{P}[A|\text{hits} \geq 1]=\frac{36+228+336}{1000-24}=\frac{75}{122}$$
Fourth question:
It is up to you...
To speed up the calculation of 3. I used the numerators previously calculated, which are the favourable cases with 1 and 2 hits, respectively.
I used integer numbers because I do not like decimals...