Expectation problem

arpitm08

"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.

alan2

They add up to one, you should recheck that. Your work is correct. Sometimes intuition fails us in probability. How can you draw three 2's? 8 choices for the first draw, 7 for the next, 6 for the next, for 8*7*6 ways. How can you draw two 2's and a 5? Well, you can draw them in 3 different orders (225), (252), (522). So you can draw them in 3*8*7*2 ways. So, as counter-intuitive as it may seem, p(6)=p(9).

arpitm08

Ahh, thank you. I knew I wasn't seeing something obvious. It just seemed like it couldn't be the same. Also, I didn't calculate 2 C 2 correctly. I didn't realize it was 1, not 2 like I was thinking for some reason.

awkward

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 [itex]X_i[/itex]. It should be clear that [tex]E[X_i] = 26/10[/tex] for i = 1,2,3. So [tex]E[X_1 + X_2 + X_3] = E[X_1] + E[X_2] + E[X_3] = 3 \times 26/10[/tex]
Here we have used the theorem [itex]E[X+Y] = E[X] + E[Y][/itex]. 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 [itex]X_i[/itex]'s are not independent.