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?

1/7

Step-by-step explanation:

The following information is missing:

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.

Let's call:

A: captain hits, P (A) = 1/2

A': captain miss, P (A') = 1 - P (A) = 1/2

B: pirate hits

We want to calculate P (A'∩B). This can be computed as follows:

P (A'∩B) = P (A') * P (B|A')

where P (B|A') means pirate hits given that captain miss. We know that P (B|A') = 2/7. Replacing data into the equation:

P (A'∩B) = (1/2) * (2/7) = 1/7