Captain Daniel has a ship, the H. M. S Crimson Lynx. The ship is two furlongs from the dread pirate Stephanie and her merciless band of thieves. If his ship hasn't already been hit, Captain Daniel has probability / dfrac{3}{4} 4 3 start fraction, 3, divided by, 4, end fraction of hitting the pirate ship. If his ship has been hit, Captain Daniel will always miss. If her ship hasn't already been hit, dread pirate Stephanie has probability / dfrac{1}{7} 7 1 start fraction, 1, divided by, 7, end fraction of hitting the Captain's ship. If her ship has been hit, dread pirate Stephanie will always miss. If the Captain and the pirate each shoot once, and the pirate shoots first, what is the probability that both the pirate and the Captain hit each other's ships?

Question

Caden Thomas

Mathematics

29 January, 17:19

Captain Daniel has a ship, the H. M. S Crimson Lynx. The ship is two furlongs from the dread pirate Stephanie and her merciless band of thieves. If his ship hasn't already been hit, Captain Daniel has probability / dfrac{3}{4} 4 3 start fraction, 3, divided by, 4, end fraction of hitting the pirate ship. If his ship has been hit, Captain Daniel will always miss. If her ship hasn't already been hit, dread pirate Stephanie has probability / dfrac{1}{7} 7 1 start fraction, 1, divided by, 7, end fraction of hitting the Captain's ship. If her ship has been hit, dread pirate Stephanie will always miss. If the Captain and the pirate each shoot once, and the pirate shoots first, what is the probability that both the pirate and the Captain hit each other's ships?

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Answers (1)

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Weston Juarez · Answer 1

Zero (0)

Step-by-step explanation:

From the given information, we are to determine the probability that both ships are hit given that the pirate shoots first.

There are two scenarios

Since the pirate is to shoot first

If the pirate misses, the Captain's ship is not hit. Therefore there's no chance of both ships getting hit. If the pirate hits, we are told that if his ship has been hit, Captain Daniel will always miss. Therefore there is also no chance of both ships getting hit.

P (both ships to be hit given that pirate shoots first) = 0