
Stats, Seems so simple
10 people get on the elevator from the ground floor. They are equally likely to get down at any of the 10 floors in the building, and choose their floors independently.
a)Determine the expected number of floors where no one gets down from the elevator.
b)How would your answer change if the 10 people get down in pairs?
Me and my friend are stuck on this question he thinks the answer to part a) is 1.
and the answer to part b is 0.5. But for part b) i think the answer has to be more the 5 anyway so i think he is wrong.
I would apprecate any help as i dont even know which distribution to use.

For any floor, the chance on x number of people going there can be modelled with a binomial distribution of Bin(10, 1/10) (the same as throwing a tensided dice once for every person).
So the chance of a floor having 0 people is
$\displaystyle P(X=0) = \left( \frac{10!}{0!10!} \right) (1/10)^0 (9/10)^{10} = (9/10)^{10} \approx 0.35$
The chance of x number of floors having 0 people is again a binomial Bin(10, 0.35) of which the expected value is 10 * 0.35 = 3.5
For (b) I assume you mean there are 5 people in the elevator.
This changes the initial binomial to Bin(5, 1/10) for which
$\displaystyle P(X=0) = (9/10)^5 \approx 0.59$
The chance of x number of floors with 0 people becomes Bin(10, 0.59). The expected value of that is 10 * 0.59 = 5.9