Question

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?

Answer 1

Answer:

18/35 on my khan

Step-by-step explanation:

Answer 2

Answer:

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