Please help me with this problem...
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?

The probability that they both miss their targets is 1/3

Step-by-step explanation:

In this question, we are trying to calculate the probability that if the captain and the pirate fires the canon in their ships at the same time, they both miss.

We proceed as follows;

Let the probability of the captain hitting the pirate be P(c) = 1/2

The probability of the captain missing is thus ; 1 - 1/2 = 1/2 or 0.5

Let the probability of the pirate hitting the captain be P(r) = 1/3 , the probability of him not hitting the captain is 1-1/3 = 2/3

Now the probability of both of them missing is = probability of the captain missing a fired shot * probability of the pirate missing a fired shot

Hence, we have; 2/3 * 1/2 = 2/6 = 1/3

the expected value of the random variable : 0, 1 and 2

step-by-step explanation:

168

step-by-step explanation:

i am pretty sure that the answer is 168