Expectation problem


by arpitm08
Tags: expectation
arpitm08
arpitm08 is offline
#1
Feb28-12, 10:30 PM
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.
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alan2
alan2 is offline
#2
Feb28-12, 10:59 PM
P: 188
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
arpitm08 is offline
#3
Feb29-12, 12:36 AM
P: 48
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
awkward is offline
#4
Feb29-12, 05:08 PM
P: 325

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 [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.


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