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

  • Thread starter Thread starter arpitm08
  • Start date Start date
  • Tags Tags
    Expectation
AI Thread Summary
The discussion revolves around calculating the expected value of three randomly chosen chips from a bowl containing 8 chips worth $2 each and 2 chips worth $5 each. Participants clarify the probabilities for possible outcomes: drawing three $2 chips (totaling $6), two $2 chips and one $5 chip (totaling $9), and two $5 chips and one $2 chip (totaling $12). There is a realization that the probabilities for drawing $6 and $9 are equal, which initially seemed counterintuitive. Additionally, an alternative method for calculating the expected value using the linearity of expectation is suggested, simplifying the process. The conversation emphasizes the importance of correctly calculating combinations and understanding probability principles.
arpitm08
Messages
50
Reaction score
0
"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.
 
Physics news on Phys.org
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).
 
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.
 
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.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

Similar threads

Replies
8
Views
2K
Replies
18
Views
1K
Replies
4
Views
1K
Replies
1
Views
1K
Replies
3
Views
4K
Replies
5
Views
1K
Replies
9
Views
2K
Back
Top