Jennifer and Frood, according to me both of you are saying almost the same thing in a slightly different ways.
In regards to the stats of the AA gun, there is a properties that may be used: the theorem of total probability.
If we consider A the event that first dice hit and B the event that second dice hit, we all agree that each probability a priori is P(A) = P(B) = 1/6 = 0.166666 approximately 17%.
The probability of the two events, that are independent, is:
P(A AND B) = P(A)*P(B) = 1/36 = 0.027777 approximately 3%.
Given the fact that the two events are independent but not disjoint (or they are not mutually esclusive, that is the same) the probability of hitting at least one of the two aircraft is:
P(A OR B) = P(A)*P(B) - P(A AND B) = (1/6) * (1/6) - 1/36 = 11/36 = 0.305555 approximately 31%. (as Frood said)
There is another way to compute the same probability, considering the complementary event A’ and B’, that are the events in which aircraft is not hit.
The a priori probability in this case is 5/6. So:
P(A OR B) = 1 - P(A’ AND B’) = 1 - (5/6) * (5/6) = 1 - 25/36 = 11/36 = 0.305555
What Jennifer calculated is the conditional probability of P(A) given that P(B) is not verified. Being the events independent we have that P(A AND B) = P(A)*P(B) and so, for the Bayes Law:
P(A|B) = P(A AND B)/P(B) = P(A)*P(B)/P(B) = P(A) = 0.166666
So you are looking different aspects of the problem but you have calculated the correct probabilities!