Question

Captain Umaima has a ship, the H.M.S. Khan. The ship is two furlongs from the dread pirate Nadia and her merciless band of thieves. The Captain has probability 3/7 of hitting the pirate ship. The pirate only has one good eye, so she hits the Captain's ship with probability 2/5. If both fire their cannons at the same time, what is the probability that the Captain hits the pirate ship, but the pirate misses.

Answer 1

Answer:

$\frac{9}{35}$.

Step-by-step explanation:

We have been given that captain Umaima has probability 3/7 of hitting the pirate ship.

The pirate only has one good eye, so she hits the Captain's ship with probability 2/5.

The probability that pirate misses the captain's ship will be:

$\text{Probability that pirate misses captain's ship}=\text{1-probability that pirate hits captain's ship}$    

$\text{Probability that pirate misses captain's ship}=1-\frac{2}{5}$  

$\text{Probability that pirate misses captain's ship}=\frac{5}{5}-\frac{2}{5}$

$\text{Probability that pirate misses captain's ship}=\frac{5-2}{5}$    

$\text{Probability that pirate misses captain's ship}=\frac{3}{5}$  

So the probability that pirate misses captain's ship will be 3/5.

Since both events are independent, so by the multiplication rule of probability we will get,

$\text{Probability that captain hits the pirate ship, but pirate misses captain's ship}=\frac{3}{7}\times \frac{3}{5}$

$\text{Probability that captain hits the pirate ship, but pirate misses captain's ship}=\frac{3*3}{7*5}$

$\text{Probability that captain hits the pirate ship, but pirate misses captain's ship}=\frac{9}{35}$

Therefore, the probability that the Captain hits the pirate ship, but the pirate misses will be $\frac{9}{35}$.