- #1
Jane_S
- 3
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Homework Statement
Hi, I'm considering the following problem: there are x males and y females, each of whom gets on an elevator at the bottom floor. There are 2k floors above this one (so 2k total stops). Furthermore, men are twice as likely to get off at odd floors as compared to even ones, and women are twice as likely to get off at even floors as compared to odd ones. People get off independently. If we let W equal the number of stops the elevator makes, find E(W) and Var(W). I'm not too worried about actually finding the variance and the expected value, I just want to see if I'm correct about the probability that someone gets off.
Homework Equations
Definitely want to use indicator random variables, and also probably the binomial theorem.
The Attempt at a Solution
So I understand that if people didn't care about odd/even floors, probability that passenger i gets off at a given floor is 1/(2k). In order to find the probability that we have a stop at floor i, we would do 1-((x+y)choose 0) * (1/(2k))^0*(1-(1/(2k)))^(m+n),
just subtracting from one the chance that nobody gets off at floor i.
The problem I'm running into is taking into account the odd/even part. I know the chance that someone gets off on their preferred type of floor is 2/2k, and the chance they get off on their other one is 1/2k. Since the chance that any given floor is your preferred type is .5, i get .5(2/2k*1/2k), or 3/4k. Thus, the chance that the elevator stops at a floor is 1-(1-(3/4k))^(m+n) (since the other terms disappear because of the zero). I fear that this is incorrect, however, because it doesn't deal with men and women seperately; unfortunately, I have no idea how to take this into account!
Thanks for the help in advance :)