I believe the answer to 1 is:
This is because we are conducting 8 total interviews (numerator) in two rooms that have four slots each (denominator).
2 and 3 seem like you'd use the negative binomial distribution to solve.
Four people are going for an interview between 9am and 10am. Each person is to be interviewed in two separate rooms for a total of 30 minutes. They are interviewed for 15 minutes in room 1. And they are interviewed for 15 minutes in room 2. Each person is interviewed by themselves (no doubling up occurs in a room in a specified time slot).
1) - How many ways can the interviews be scheduled during the one hour period.
2) - If a person arrives at 9am, what is the probability they will have their interviews consecutively (eg 1 at 9am, the other at 9:15.... or 1 at 9:30, the other at 9:45 etc...)
2) - If someone turns up at 9am, what is the probability they will be able to leave after the third interview since they need to attend another appointment somewhere else?
Any help with these ones will be greatly appreciated!!!
I believe the answer to 1 is:
This is because we are conducting 8 total interviews (numerator) in two rooms that have four slots each (denominator).
2 and 3 seem like you'd use the negative binomial distribution to solve.
Room 1: Slot 1,2,3,4
Room 2: Slot 1,2,3,4
You can arrange 4 people to occupy Slot's in Room 1 in 4! ways
Once done, Slots in Room 2 can give given to 4 people, with care that no one occupies the same slot in both room 1 and room 2. This just de-arrangement for n = 4 which is equal to 9.
So 4!*9 is the ans
Derangements are permutations of a set such that there is no element in it's original position. You need to use this in this problem because you can't have a person sitting in both 9:00 am timeslots etc. Google 'derangements' (not de arrangemnts) and you'll find a formula on wiki
Derangement - Wikipedia, the free encyclopedia
Plz look at this - it explains it pretty well