If the Captain and the pirate each shoot once, and the Captain shoots first, what is the probability that the Captain misses the pirate ship, but the pirate hits?

Question

Captain Nadia has a ship, the H.M.S Crimson Lynx. The ship is two furlongs from the dread pirate Tiffany and her merciless band of thieves.

If her ship hasn't already been hit, Captain Nadia has probability 1/2 of hitting the pirate ship. If her ship has been hit, Captain Nadia will always miss.

If her ship hasn't already been hit, dread pirate Tiffany has probability 2/7 of hitting the Captain's ship. If her ship has been hit, dread pirate Tiffany will always miss.

Answer 1

Let A be the event that the captain hits, A' is the complement (captain misses). Let B be the event that the pirate hits, B' is the complement (pirate misses).

So, written as a statement, what the question is asking for is: P(A' ∩ B). We're told that the captain shoots first, and they each shoot once. Conditional probability tells us that the probability of an intersection can be found by multiplying the probability of the first event, times the probability of the second event given the first occurred: P(A' ∩ B) = P(A')*P(B|A').

P(A') is the probability that the captain misses, which is 1/2 (since the captain shoots first, her ship can not have been hit yet).

P(B|A') is the probability that the pirate hits, given that the captain misses, which is 2/7.

So, P(A' ∩ B) = (1/2)*(2/7) = 1/7