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

\(\displaystyle{E}{\left({Y}\right)}={1.1}{E}{\left(\chi_{{{\left({X}\in{\left\lbrace{0},{2}\right\rbrace}\right)}}}\right)}-{E}{\left(\chi_{{{\left({X}={1}\right)}}}\right)}\)

\(\displaystyle={1.1}{P}{\left({X}={0}\right)}+{1.1}{P}{\left({X}={2}\right)}-{P}{\left({X}={1}\right)}\)

\(\displaystyle={2}\cdot{1.1}\cdot{\frac{{{10}}}{{{45}}}}-{\frac{{{25}}}{{{45}}}}=-{\frac{{{1}}}{{{15}}}}\)