# A box contains 5 red and 5 blue marbles. Two marbles are withdrawn randomly. If

A box contains 5 red and 5 blue marbles. Two marbles are withdrawn randomly. If they are the same color, then you win 1.10; if they are different colors, then you win -1.10; if they are different colors ,then you win−1.00. (This is, you lose \$1.00.) Calculate (a) the expected value of the amount you win; (b) the variance of the amount you win.

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SaurbHurbkj

Define random variable X that marks the number of red marbles that have been drawn. We have that $$\displaystyle{X}\in{\left\lbrace{0},{1},{2}\right\rbrace}$$ and by the nature of the problem, we have that X has Hypergeometric distribution, i.e.
$$P\left(X = i\right) = \frac{\left(\begin{array}{c}5\\ i\end{array}\right)\left(\begin{array}{c}5\\ 2-i\end{array}\right) }{\left(\begin{array}{c}10\\ 2\end{array}\right)}$$
Now, consider random variable Y that marks the amount of money that we win. It can be written as function of X as follows
$$\displaystyle{Y}={1.1}\cdot\chi_{{{\left({X}\in{\left\lbrace{0},{2}\right\rbrace}\right)}}}-{1}\cdot\chi_{{{\left({X}={1}\right)}}}\rbrace$$
(a)
Hence, using the linearity of expectation, we have that
$$\displaystyle{E}{\left({Y}\right)}={1.1}{E}{\left(\chi_{{{\left({X}\in{\left\lbrace{0},{2}\right\rbrace}\right)}}}\right)}-{E}{\left(\chi_{{{\left({X}={1}\right)}}}\right)}$$
$$\displaystyle={1.1}{P}{\left({X}={0}\right)}+{1.1}{P}{\left({X}={2}\right)}-{P}{\left({X}={1}\right)}$$
$$\displaystyle={2}\cdot{1.1}\cdot{\frac{{{10}}}{{{45}}}}-{\frac{{{25}}}{{{45}}}}=-{\frac{{{1}}}{{{15}}}}$$

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(b)
Now, let's deal with the variance. We will use that $$\displaystyle{V}{a}{r}{\left({Y}\right)}={E}{\left({Y}^{{{2}}}\right)}-{E}{\left({Y}\right)}^{{{2}}}$$. Observe that
$$\displaystyle{Y}^{{{2}}}={1.1}^{{{2}}}\cdot\chi_{{{\left({X}\in{\left\lbrace{0},{2}\right\rbrace}\right)}}}+{1}^{{{2}}}\cdot\chi_{{{\left({X}={1}\right)}}}$$
so similarly
$$\displaystyle{E}{\left({Y}^{{{2}}}\right)}={1.21}{P}{\left({X}={0}\right)}+{1.21}{P}{\left({X}={2}\right)}+{P}{\left({X}={1}\right)}={\frac{{{82}}}{{{75}}}}$$
Finally, we have that
$$\displaystyle{V}{a}{r}{\left({Y}\right)}={\frac{{{82}}}{{{75}}}}-{\frac{{{1}}}{{{225}}}}={\frac{{{49}}}{{{45}}}}$$