Captain Ben has a ship, the H. M. S Crimson Lynx. The ship is five furlongs from the dread pirate Luis and his merciless band of thieves. If his ship hasn't already been hit, Captain Ben has probability / dfrac{2}{3} 3 2 start fraction, 2, divided by, 3, end fraction of hitting the pirate ship. If his ship has been hit, Captain Ben will always miss. If his ship hasn't already been hit, dread pirate Luis has probability / dfrac{1}{3} 3 1 start fraction, 1, divided by, 3, end fraction of hitting the Captain's ship. If his ship has been hit, dread pirate Luis will always miss. If the Captain and the pirate each shoot once, and the pirate shoots first, what is the probability that the pirate hits the Captain's ship, but the Captain misses?

We can see that this is dependent probability. We can find dependent probability of happening event A then event B by multiplying probability of event A by probability of event B given that event A already happened.

Step-by-step explanation:

In our case event A is pirate hitting captain's ship and event B is captain missing pirate's ship. We have been given that pirate shoots first so pirate's ship can't be hit before pirate shoots his cannons. So probability of hitting captain's ship is 1/3. We have been given that if Captain Ben's ship is already hit then Captain Ben will always miss. So the probability of Captain missing the dread pirate's ship given the pirate Luis hitting the Captain ship is 1. Now to find probability that pirate hits Captain, but Captain misses we will multiply our both probabilities.