Calculate Expectation Problem for 3 Randomly Chosen Chips from a Bowl of 10

[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 receive 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 $X_i$. 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.

 

[blue_raver22]

It appears that you have made a mistake in your calculations for p(9) and p(12). The correct formulas should be:

p(9) = (8 C 2)(2 C 1)/(10 C 3)
p(12) = (8 C 1)(2 C 2)/(10 C 3)

The reason for this is that for p(9), we are choosing 2 chips from the 8 $2 chips and 1 chip from the 2 $5 chips. This is represented by (8 C 2)(2 C 1). Similarly, for p(12), we are choosing 1 chip from the 8 $2 chips and 2 chips from the 2 $5 chips, which is represented by (8 C 1)(2 C 2).

Once these probabilities are calculated correctly, you can proceed with finding the expectation. This is done by multiplying each possible value by its corresponding probability and then adding all the products together. In this case, the expectation would be:

E(X) = 6*p(6) + 9*p(9) + 12*p(12)

I hope this helps clarify your calculations. Remember to always double check your formulas and calculations to avoid any errors.