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

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Discussion Overview

The discussion revolves around calculating the expected value when randomly selecting 3 chips from a bowl containing 10 chips, with 8 chips valued at $2 and 2 chips valued at $5. Participants explore different methods for determining the probabilities associated with the possible sums of the selected chips.

Discussion Character

  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • One participant outlines their approach to calculating the expectation by identifying possible values for the sum of the chips and attempting to compute the associated probabilities.
  • Another participant suggests that the initial calculations for probabilities might be incorrect and provides a detailed explanation of how to derive the probabilities for drawing the chips, emphasizing the importance of considering different orders of drawing.
  • A later reply acknowledges a misunderstanding regarding the calculation of combinations and expresses gratitude for the clarification provided by the second participant.
  • Another participant introduces an alternative method for calculating the expected value using the linearity of expectation, suggesting that the expected value can be computed directly without considering the dependence of the draws.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correctness of the initial probability calculations. While one participant believes the calculations are incorrect, another asserts they are correct. The discussion remains unresolved regarding the initial approach, but there is agreement on the alternative method presented.

Contextual Notes

Some participants express uncertainty regarding the calculations of combinations and the implications of drawing chips without replacement. The discussion highlights the complexity of probability calculations in this context.

arpitm08
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"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.
 
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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.
 

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