How to find the probability that a group of people is not allowed to enter a country given some factors? I have this scenario:
There are two countries A and B.
A sends a group of N people to B...
There is an independent probability of p that a person gets fatigued while traveling to B.
And I have these questions:


*

*What is the probability that 0 people in the group are fatigued
by the time they arrive to B?

*What is the probability that m people are fatigued by the time
they get to B?

*What is the probability that at most m people are fatigued by the
time they get to B?

*Consider the case where B rejects the whole group from entering the country if more than m people are fatigued by the time they get there. In which case, the group returns to A and A will resend the people all over again... What is the probability that A has to send the group of people k times before it is accepted into the country?

*Given those probabilities, what is the expected number of times A is supposed to send the group of N people to B before it gets accepted?
My confusion arises because I am unsure of how to approach this...
For the first part 1, I assumed a simple binomial probability... thus the probability of m = 0 is $(1-p)^{n}$. 
For part 2, I generalized part 1:
Probability of m being fatigued = $P_{fat}(m) = {N \choose m} p^{m}(1-p)^{N-m}$
For part 3, I assumed it is just an aggregated version of part 2:
Probability of m being fatigued = $P_{atMostFat}(m) = \sum_{k = 0}^{m} P_{fat}(k) $
Unfortunately, I am not sure on how to approach parts 4 and 5. And I am not even sure if what I did for 1,2,3 is valid.
For 4 I tried to find $1 - P_{atMostFat}(m)$ because IMO that is the probability that more than m are fatigued... But I am unsure on how to relate it to k...
Can you help me please?
 A: Your answers for parts 1-3 look good to me. We use the binomial distribution since we're doing a fixed number of trials (or sending a fixed number of people independently of one another) where there's a fixed probability of success (i.e. not getting fatigued, so 1-p) or failure (i.e. getting fatigued, probability p) for each trial.
For part 4, you have the probability (from part 3) that at most m people are fatigued, and as you pointed out, you can relate this to the probability that more than m people are fatigued. Now, we're considering trials (sending a group of people independently of the past) where there's a fixed probability of a success (i.e. m or fewer people are fatigued), or failure (i.e. more than m people are fatigued) for each trial.
This is very similar to what's described above, but there's a key difference. In the first scenario, we had a fixed number of 'trials' (i.e. we sent a fixed number of people). In part 4, we aren't sending the group a fixed number of times, as it will depend on when the first success occurs (e.g. if fewer than m people were fatigued in the first journey, the group would be let into B and there would be no more journeys, if this didn't happen, there would be more). Do you know what type of distribution we will want to look at in this scenario?

 Geometric distribution

For part 5, this is asking what the expected value of the case we were considering in part 4 is, which is a standard result.

 The expected number of trials (or times we send the group) is 1/q, where q is the probability of the trial being a success, which I'll leave you to figure out.

I hope that helps.
