Captain Jessica has a ship, the H.M.S Crimson Lynx. The ship is two furlongs from the dread pirate Ishaan and his merciless band of thieves. If her ship hasn't already been hit, Captain Jessica has probability \dfrac{2}{5} 5 2 start fraction, 2, divided by, 5, end fraction of hitting the pirate ship. If her ship has been hit, Captain Jessica will always miss. If his ship hasn't already been hit, dread pirate Ishaan has probability \dfrac{3}{8} 8 3 start fraction, 3, divided by, 8, end fraction of hitting the Captain's ship. If his ship has been hit, dread pirate Ishaan will always miss. 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?

Captain Jessica shoots first. This means her ship has not already been hit, which means she has a 2/5 chance of hitting the pirate ship. This also means she has a 1-2/5 = 3/5 chance of missing the pirate ship.

Since Captain Jessica misses, dread pirate Ishaan has a 3/8 chance of hitting the captain's ship. This means he has a

3/5(3/8) = 9/40 chance of the captain missing the pirate and the pirate hitting the captain.

ap8997154 ap8997154 · Answer 2 · 2022-04-20T15:17:24+02:00

The probability tells the chances of an event occurring. The probability that the Captain misses the pirate ship, but the pirate hits are 9/40.

What is Probability?

The probability helps us to know the chances of an event occurring.

[tex]\rm Probability=\dfrac{Desired\ Outcomes}{Total\ Number\ of\ outcomes\ possible}[/tex]

As it is given that the probability of the captain hitting the pirate ship is 2/5, therefore, the probability of the captain not hitting the pirate ship will be 3/5 (1- 2/5).

Also, it is mentioned that the probability of the pirate hitting the captain is 3/8, therefore, the probability of the pirate not hitting the captain can be written as 5/8.

We need to calculate the probability of the captain's ship shooting first but not hitting the pirate because in this case, only the pirate can hit back at the captain. Also, it is needed that the two events occur one after the other. Therefore, the probability that the Captain misses the pirate ship, but the pirate hits will be

[tex]P = (\text{Probability captain don't hit pirate}) \times (\text{Probability the pirate hit captain})\\\\\\P = \dfrac{3}{5} \times \dfrac{3}{8}= \dfrac{9}{40}[/tex]

Hence, the probability that the Captain misses the pirate ship, but the pirate hits is 9/40.

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