Q.4.35 - A First Course in Probability

Question

A box contains $5$ red and $5$ blue marbles. Two marbles are withdrawn randomly. If they are the same color, then you win $\backslash (1.10$; if they are different colors, then you win $-\backslash )1.00$. (That is, you lose $\$1.00$.) Calculate

(a) the expected value of the amount you win;

(b) the variance of the amount you win.

Step-by-Step Solution

Answer

(a)  The expected value of the amount you win will be  $-0.0667$.

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

Step 1: Given information (Part a)

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 Two marbles are withdrawn randomly.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 $-\mathrm{\$}1.00$.

Step 2: Solution (Part b)

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 $=\frac{4}{9}$

$\mathrm{X}=-1.0$

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

$=\frac{5}{9}$

$E\left(X\right)=1.1\times \frac{4}{9}-1.0\times \frac{5}{9}$

$=-0.0667$

Step 3: Final answer (Part a)

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

Step 4: Given information (Part b)

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 Two marbles are withdrawn randomly.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

Step 5: Solution (Part b)

Let the calculation will be,

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

$=0.537778+0.555556$

$=1.0933$

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

$=1.0933-(-0.0667{)}^2$

$=1.089$

Step 6: Final answer (Part b)

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