Register to reply 
Expectation problemby arpitm08
Tags: expectation 
Share this thread: 
#1
Feb2812, 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. 


#2
Feb2812, 10:59 PM

P: 199

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 counterintuitive as it may seem, p(6)=p(9).



#3
Feb2912, 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.



#4
Feb2912, 05:08 PM

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 [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. 


Register to reply 
Related Discussions  
Express moment / expectation value in lower order expectation values  General Math  3  
Expectation Values/Braket Problem  Advanced Physics Homework  5  
Expectation value problem  Advanced Physics Homework  1  
The Expectation of X and the Expectation of X squared (discrete math)  Calculus & Beyond Homework  2  
Expectation value problem  Calculus & Beyond Homework  5 