The problem reads like this: The number of calls the company receives is 4 per hour. a) Find probability to get exactly 5 between 09:00 and 12:00 of a given day
b) Given that there the company has received at least 6 between 12:00-14:00, what is the probability to get 8 calls for the same period?
c) For a week of 7 days, what is the probability of having 4 days when there are not phone calls between 12:30-13:00?
For a) I already know $P(x=5, \lambda=2\cdot4=8 ) = \frac{8^5 e^{-8}}{5!}$.
However, I am not sure what needs to be done for b) and c)... For b) I guess needs something like
$P(x\ge 8) = 1- P(x=6) - P(x=7)$ ?
But then, it does not give any clue about the numerical values of $P(x=6), P(x=7)$ so I conclude the given information has to be combined in another way.
And c, is a complete mystery, so any hint would be much appreciated!.
For b, you want the conditional probability $$P(X=8|X\geq 6) = \frac{P(X=8 \cap X \geq 6)}{P(X\geq 6)}= \frac{P(X=8)}{P(X\geq 6)}$$
For c, you want to find the probability of zero calls between $12:30$ and $13:00$ on any day, and then use a binomial distribution. You have $7$ days, and you want any $4$ of them to have zero.
Remember, the requirements for a binomial is that there are a fixed number of trials, the trials are independent, there are only two outcomes, and the probability is the same every trial. How many times are you trying to have no calls, and how many times do you?