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Given 3 red balls of value 3, 2 blue balls of value 5, and 1 yellow ball of value 25

a) What is the expectation for X of drawing two balls? Where X is the value.

My issue here is that we have no way to be sure what ball was drawn the first time, I have calculated the X of drawing one ball to be 7.33.. So at first I thought I might subtract this off my answer to simulate that much value being "lost" because it couldn't be drawn again. But that doesn't make much sense, so I am not sure how to proceed.

b) Suppose you are allowed to keep drawing balls until you draw the yellow ball. (So you will draw at most six times, since the balls are not replaced after each draw.) What is the expectation for X of balls drawn?

I know how to do this problem in regards to dice, but again, the lack of replacement is giving me trouble.

Thanks.

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2 Answers 2

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a) Let $X_1, X_2$ be the values of the first and second balls, respectively.   Then by the Linearity of Expectation we know that: $\;\mathsf E(X_1+X_2) = \mathsf E(X_1)+\mathsf E(X_2)$ .   L.o.E. does not require the random variables to be independent.

The LHS is $\mathsf E(X)$ and the RHS is easy to calculate, as lack of replacement is not an issue.   $\mathsf E(X_1)=\mathsf E(X_2) = \frac{3 \times 3+ 2 \times 5 + 1 \times 25}{3+2+1}$

$$\therefore \mathsf E(X_1+X_2) = \frac{44}{3}$$


If you have doubts about the Linearity of Expectation (it is a wee bit counter intuitive), simply evaluate the hypergeometrically distributed probabilities and calculate the expectation from those:

$$\begin{align}\mathsf E(X) & = (3+3)\frac{\binom{3}{2}}{\binom{6}{2}}+(3+5)\frac{\binom{3}{1}\binom{2}{1}}{\binom{6}{2}}+(3+25)\frac{\binom{3}{1}\binom{1}{1}}{\binom{6}{2}} +(5+5)\frac{\binom{2}{2}}{\binom{6}{2}}+(5+25)\frac{\binom{2}{1}\binom{1}{1}}{\binom{6}{2}} \\[1ex] & = \dfrac{6\times 3+8\times 6+28\times 3+10\times 1+30\times 2}{15} \\[1ex] & = \dfrac{44}{3} \end{align}$$


b) Let $N$ be the count of draws until a yellow ball is drawn.   Can you calculate $\mathsf E(X\mid N{=}n)$, and $\mathsf P(N{=}n)$?

$\mathsf E(X\mid N=n)=\sum_{i=1}^{n-1}\mathsf E(X_i\mid N{=}n) = (n-1)\cdot\frac{3 \times 3+ 2 \times 5}{3+2}+25$

Then by the Law of Iterated Expectation $\mathsf E(X)=\mathsf E(\mathsf E(X\mid N))$

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  • $\begingroup$ Thanks a bunch, this makes sense. $\endgroup$
    – Luke
    Dec 3, 2015 at 4:34
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Suppose you are allowed to keep drawing balls until you draw the yellow ball. (So you will draw at most six times, since the balls are not replaced after eachdraw.) What is the expectation for X of balls drawn?

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