To maximise their chances the duellists prefer to be left with a weaker opponent. So Bob would not shoot at Alice in preference to Carol, and Carol will not shoot at Alice in preference to Bob. Therefore Alice will not be shot at until Bob or Carol is dead and she will either be left standing with Bob or Carol, with or without the shot.
Probability of Alice, with shot, surviving against Bob is given by:
p = Pr(A hits B) + Pr(A misses B) * Pr (B misses A) * p
p = 1/3 + 2/3 * 1/3 * p
p = 3/7
Probability of Alice, without shot, surviving against Bob is given by:
p = Pr(B misses A) * (Pr(A hits B) + Pr (A misses B) * p)
p = 1/3 * (1/3 + 2/3 * p)
p = 1/7
Probability of Alice, with shot, surviving against Carol is given by:
p = Pr(A hits C) + Pr(A misses C) * Pr (C misses A) * p
p = 1/3 + 2/3 * 0 * p
p = 1/3
Probability of Alice, without shot, surviving against Carol is given by:
p = Pr(C misses A) * (Pr(A hits C) + Pr (A misses C) * p)
p = 0 * (1/3 + 2/3 * p)
p = 0
So, her probability of surviving from each position is:
Bob, with shot: 3/7
Carol, with shot: 1/3
Bob, without shot: 1/7
Carol, without shot: 0
So Alice is best off not killing anyone since the advantage she gains by having the first shot exceeds any possible benefit of facing Bob rather than Carol. She should shoot into the air.
Given that Alice is neither going to shoot at them, or be shot at by them until one is dead, Bob and Carol are essentially in a two person duel, the winner to face Alice. They cannot improve their chances by forgoing a shot, so they shoot. Bob wins that 2/3 of the time, Carol 1/3.
Alice wins 2/3 * 3/7 + 1/3 * 1/3 = 25/63.
Bob wins 2/3 * 4/7 = 24/63.
Carol wins 1/3 * 2/3 = 14/63.
Random variable $X$ is the number of shots needed to get the shot where Mark doesn't miss. The distribution if $X$ is geometric. You noted that
the number of shots can only be 1 or 2
So $Pr(X=0)=0$, as well as $Pr(X=-5)$ or $\Pr(X=-150)$. Do we need to take the last two probabilities into account?
The other way to find the probability that Mark will need to take at least 3 shots until he has a one where he doesn't miss is to note that he should have two shots where he misses. Really: if he misses in first two shots then he needs in third shot. And vice versa: if at least three shots are required, he miss in first two shots.
Therefore the required probability is $0.15^2=0.0225$.
Best Answer
You can use the binomial distribution instead with parameters $n=10$ and $p=10\%$. Specifically, the number of misses $Y$ in the first $n=10$ shots is binomially distributed with parameters $n=10$ and $p$ as above, i.e. $p=10\%$. Then the probability that the second miss comes no later than the $10$-th shot is equal to the probability that there are at least two misses in the first $10$ shots. Hence you want to calculate \begin{align}P(Y\ge 2)&=1-P(Y<2)\\[0.2cm]&=1-P(Y=0)-P(Y=1)\\[0.2cm]&=1-\dbinom{10}{0}(0.10)^0(1-0.10)^{10}-\dbinom{10}{1}(0.10)^1(1-0.10)^{10-1}\end{align}