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 Two marbles are withdrawn randomly.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

aidmoon2x 2021-11-21 Answered
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 Two marbles are withdrawn randomly.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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Expert Answer

SaurbHurbkj
Answered 2021-11-22 Author has 1071 answers

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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Eprint
Answered 2021-11-23 Author has 447 answers
(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
\(\displaystyle{V}{a}{r}{\left({Y}\right)}={\frac{{{82}}}{{{75}}}}-{\frac{{{1}}}{{{225}}}}={\frac{{{49}}}{{{45}}}}\)
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