A and B are involved in a duel. The rules of the duel are that they are allowed to pick up their guns simultaneously. If one or both are hit, the duel ends. If both shots miss, then they repeat the process. Suppose that the results of the shots are independent and that each shot of A will hit B with prob $p_A$ and each shot B will hit A with prob $p_B$.
Compute: a) The probability that A is not hit. This is just $1-p_B$ for each trial. So if we assume there are $n$ duels then B has to be hit. Summing over the $n$ duels gives the probability as $ \sum_{i=1}^{n} (1-p_B)^{n-i} p_A$
b) The probability that both duelists are hit. I said this would be $p_A p_B$ for any one duel
c) the prob that the duel ends after the nth round of shots. It can end in 3 ways: A hits B, B hits A or they both hit each other. So I believe we want $(1-p_A)^{n-1}p_A + (1-p_B)^{n-1}p_B + (1-p_Ap_B)^{n-1}p_Ap_B$
d) The cond. prob that the duel ends after the nth round of shots given that A is not hit. So using the def of cond. prob I get $(1-p_B)^{n-1}p_A/\sum_{i=1}^{n} (1-p_B)^{n-i} p_A$, where P(duel ends after nth round and A not hit) is numerator and P(A not hit) as in a)
e) the cond, prob that the duel ends after the nth round given that both duelists are hit. I used Bayes to write this as P(A and B both hit| duel ends on nth round)P(duel ends on nth round)/P(A and B both hit) I said the first term in the numerator was 1/3. If the duel ends on the nth trial, then either A hit B, B hit A or they both hit each other => 1/3. Then I just substituted what we had already in the previous parts.
I don't have the answers to these questions. Can someone make any comments about them? Many thanks