Finding how 3 hunter can share their prey

Question

Three hunters hit the moving target with probabilities $0,6, 0,4$ and $0,2$. When they saw a deer they shot in one time. After it they saw that only one bullet reached the target. So how they can shere a prey?

I think first I need to find $\mathbb{P}(X)$ where $X$ is event that deer was shot. $\mathbb{P}(X)=\mathbb{P}( (A\cap \overline{B}\cap \overline{C}) \cup (\overline{A}\cap B \cap \overline{C}) \cup (\overline{A}\cap \overline{B} \cap C)) $ am I right? When I will find it I will make a proportion and will find how they can share a prey.

So if I am right, I have problem. I think I am doing something wrong. $\mathbb{P}(X)=\mathbb{P}( (A\cap \overline{B}\cap \overline{C}) \cup (\overline{A}\cap B \cap \overline{C}) \cup (\overline{A}\cap \overline{B} \cap C))= \mathbb{P}( (A\cap \overline{B}\cap \overline{C}) \cup (\overline{A}\cap B \cap \overline{C})) +\mathbb{P} (\overline{A}\cap \overline{B} \cap C) - \mathbb{P}( (A\cap \overline{B}\cap \overline{C}) \cup (\overline{A}\cap B \cap \overline{C}) \cap (\overline{A}\cap \overline{B} \cap C))= \mathbb{P}(A\cap \overline{B}\cap \overline{C}) + \mathbb{P}((\overline{A}\cap B \cap \overline{C}) \cap (\overline{A}\cap \overline{B} \cap C))-\mathbb{P}( (A\cap \overline{B}\cap \overline{C}) \cap (\overline{A}\cap B \cap \overline{C}) \cap (\overline{A}\cap \overline{B} \cap C))+\mathbb{P} (\overline{A}\cap \overline{B} \cap C) - \mathbb{P} (A\cap \overline{B}\cap \overline{C}) - \mathbb{P}( (\overline{A}\cap B \cap \overline{C}) \cap (\overline{A}\cap \overline{B} \cap C))+ \mathbb{P}( (A\cap \overline{B}\cap \overline{C}) \cap (\overline{A}\cap B \cap \overline{C}) \cap (\overline{A}\cap \overline{B} \cap C))= ?$ Where is my mistake and how I can continue it? Maybe there is easier way?

Answer 1

$X$ should be the event the deer was shot exactly once, because that is the information you have. You are trying to assess $P(A|X)$, the chance that $A$ hit the deer given that exactly one person hit the deer, and the same for the other two. The probability that $A$ hit and the others did not is $0.6\cdot (1-0.4) \cdot (1-0.2)=0.288$. Compute the corresponding probability for the other two and scale up to make the sum be $1$.