# Expectation problem

by arpitm08
Tags: expectation
 P: 48 "A bowl contains 10 chips, of which 8 are marked $2 each and 2 are marked$5 each. Let a person choose, at random and without replacement, 3 chips from this bowl. If a person is to recieve the sum of the resulting amounts, find his expectation." Here is my attempt: The possible values for X are 6(2,2,2), 9(2,2,5), and 12(2,5,5). So, now we have to calculate p(x) for each of these values in order to find the expectation. p(6) = (8 C 3)/(10 C 3), where a C b is a choose b. p(9) = (8 C 2)(2 C 1)/(10 C 3) p(12) = (8 C 1)(2 C 2)/(10 C 3) These don't add up to 1 however. and I'm sure that p(6) is not equal to p(9). Could someone explain to me what I'm doing wrong in calculating p(9) and p(12). I can do the rest of the problem from there. I just can't think of what I'm doing wrong for those two. Thanks.
 P: 327 Expectation problem Hi armpitm08, Although you seem to have the problem under control, I can't resist pointing out that there is an easier way. Let's say that the value of the ith chip drawn is $X_i$. It should be clear that $$E[X_i] = 26/10$$ for i = 1,2,3. So $$E[X_1 + X_2 + X_3] = E[X_1] + E[X_2] + E[X_3] = 3 \times 26/10$$ Here we have used the theorem $E[X+Y] = E[X] + E[Y]$. It's important to realize that this theorem holds even when X and Y are not independent. That's good for us here, because the $X_i$'s are not independent.