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I tried to solve the following problem for my probability theory class :

A dice is rolled $100$ times. If $X$ is the sum of the even numbers rolled, find the expected value and the variance of $X$.

My attempt:

Let $X_i$ be the number of times we rolled $i$ $\left(\text{for } i = \overline{1,6}\right)$. For example, if after 100 tries, we rolled $5$ seven times, then $X_5 = 7$.

Then $X=2X_2 + 4X_4 + 6X_6$, so by the linearity of expectation, we have:

$$ \mathbb{E}(X) = \mathbb{E}(2X_2 + 4X_4 + 6X_6) = 2\mathbb{E}(X_2)+4\mathbb{E}(X_4) + 6\mathbb{E}(X_6) $$

Also, $$ \mathbb{P}(X_i = k) = {100 \choose k} \left( \dfrac{1}{6} \right)^k \left( \dfrac{5}{6} \right)^{100-k} $$ Therefore, $X_i \sim \text{Bin}\left(100, \dfrac{1}{6} \right) $, so we have $\mathbb{E}(X_i) = \dfrac{100}{6}$ $$ \Rightarrow \mathbb{E}(X) = \dfrac{100}{6}(2+4+6) = 200. $$

To find the variance, I know that for two independent random variables $A$ and $B$, we have $$\text{Var}(A+B) = \text{Var}(A)+\text{Var}(B),$$ which would eventually solve my problem, but somehow, I can't figure out whether $X_2, X_4, X_6$ are independent or not... Or is this a wrong approach?

Thanks in advance for any help!

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$X_i$ are NOT independent, because $\sum\limits_{i=1}^6 X_i = 100$ for example.

Consider $Y_i\sim U\{1,2,3,4,5,6\}$, represent each dice, and $Z_i=1\{Y_i \text{ even}\}$.

Now, you want to calculate expectation and variance of $\sum\limits_{i=1}^{100}Y_iZ_i$

Try to prove:

  1. $Y_iZ_i$ and $Y_jZ_j$ are independent if $i\neq j$
  2. $E(Y_iZ_i)=2P(Y_i=2)+4P(Y_i=4)+6P(Y_i=6)$

For variance, use that $Var(X)=E(X^2)-E(X)^2$ and try to analyze the variable $(Y_iZ_i)^2$. What are its possible values?

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  • $\begingroup$ Thank you. I see now that my random variables are not independent. I'll try to find the variance with the new approach. But I think my calculation of the expected value is correct, isn't it? $\endgroup$
    – sicmath
    Feb 24 at 13:18
  • $\begingroup$ Yes! The expected value is correct $\endgroup$ Feb 25 at 14:35

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