91影视

A box contains $5$red and $5$blue marbles. Two marbles are withdrawn randomly. If they are the same color, then you win $\mathrm{\$}1.10$; if they are different colors, then you win$鈭�\mathrm{\$}1.00$.

Let,

$\mathrm{x}=1.1$

$p\left(1.1\right)=\frac{\left(\begin{array}{l}5\\ 2\end{array}\right)}{\left(\begin{array}{l}10\\ 2\end{array}\right)}+\frac{\left(\begin{array}{l}5\\ 2\end{array}\right)}{\left(\begin{array}{l}10\\ 2\end{array}\right)}$

So,$\mathrm{P}(1.1)$is the probability that either both are red or both are blue $=\frac{4}{9}$

$\mathrm{X}=鈭�1.0$

$\mathrm{p}(鈭�1.0)=\frac{\left(\begin{array}{l}5\\ 1\end{array}\right)\left(\begin{array}{l}5\\ 1\end{array}\right)}{\left(\begin{array}{l}10\\ 2\end{array}\right)}$

$=\frac{5}{9}$

$E\left(X\right)=1.1脳\frac{4}{9}鈭�1.0脳\frac{5}{9}$

$=鈭�0.0667$

The expected value of the amount you win will be$鈭�0.0667$

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 Two marbles are withdrawn randomly.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$-\$1.00$

Let the calculation will be,

$\mathrm{E}\left[{\mathrm{X}}^2\right]=(1.1{)}^2脳\frac{4}{9}+(鈭�1{)}^2脳\frac{5}{9}$

$=0.537778+0.555556$

$=1.0933$

$\mathrm{Var}鈦�\left(\mathrm{x}\right)=\mathrm{E}\left({\mathrm{X}}^2\right)鈭�\left[\mathrm{E}\right(\mathrm{X}){]}^2$

$=1.0933鈭�(鈭�0.0667{)}^2$

$=1.089$

The variance of the amount you win will be $1.089$.