Question

Captain Gabriela has a ship, the H.M.S. Khan. The ship is two furlongs from the dread pirate Daniel and his merciless band of thieves.

The Captain has probability 1/2 of hitting the pirate ship. The pirate only has one good eye, so he hits the Captain's ship with probability 1/3

If both fire their cannons at the same time, what is the probability that both the Captain and the pirate miss?

Answer 1

Answer:

1/3 or 33.33%

Step-by-step explanation:

If the Captain has a probability of 1/2 of hitting the pirate ship, then he has a probability of 1/2 of missing the pirate ship.

As for the pirate, if he has a probability of 1/3 of hitting the Captain's ship, then he has a probability of 2/3 of missing the pirate ship.

The probability that both miss after firing at the same time is:

$P=\frac{1}{2}*\frac{2}{3}=\frac{2}{6}\\P=\frac{1}{3}$

The probability of both missing is 1/3 or 33.33%.

Answer 2

Answer:

1/3

Step-by-step explanation:

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 both the Captain and the pirate miss, we just need to multiply together the probability that the captain misses and the probability that the pirate misses.

The probability that the Captain misses is 1- (the probability the Captain hits), which is 1 - 1/2 = 1/2

The probability that the pirate misses is 1- (probability the pirate hits), which is 1- 1/3=2/3

SO the probability that both the Captain and the pirate miss is 1/2 X 2/3 = 1/3

=1/3