Mathematics, 01.05.2021 14:00 vanessacox45

If the Captain hits the pirate ship, it won't affect whether she's also hit by the pirate's cannons (and vice-versa), because they both fired at the same time. So, these events are independent. Since they are independent, in order to get the probability that the pirate hits the Captain's ship, but the Captain misses, we just need to multiply together the probability that the captain misses and the probability that the pirate hits.

The probability that the Captain misses is 1 -1−1, minus (the probability the Captain hits), which is 1 - \dfrac{3}{4} = \dfrac{1}{4}1−
1-3/4-1/4

1, minus, start fraction, 3, divided by, 4, end fraction, equals, start fraction, 1, divided by, 4, end fraction.

The probability that the pirate hits is 1/4

start fraction, 1, divided by, 4, end fraction.
So, the probability that the pirate hits the Captain's ship, but the Captain misses is \dfrac{1}{4} \cdot \dfrac{1}{4} = \dfrac{1}{16}
4
1

Answer =
1/16

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If the Captain hits the pirate ship, it won't affect whether she's also hit by the pirate's cannons...

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