Question

Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the probability distribution of the number of diamond cards drawn. Also find the mean and the variance of the distribution.

Answer 1

Let X denote the random variable for the diamonds cards $\therefore X=0,1,2.$

Here $n=2,p=\frac{1}{4},q=1-p=\frac{3}{4}.$

$\begin{array}{ccccc}\text{X}& 0& 1& 2& \text{Total}\\ P\left(X\right)& 2_{C{}_0}{\left(\frac{1}{4}\right)}^{\circ }{\left(\frac{3}{4}\right)}^2=\frac{9}{16}& 2_{C{}_1}{\left(\frac{1}{4}\right)}^1{\left(\frac{3}{4}\right)}^{2-1}=\frac{6}{16}& 2_{C{}_2}{\left(\frac{1}{4}\right)}^2{\left(\frac{3}{4}\right)}^{2-2}=\frac{1}{16}& \\ XP\left(X\right)& 0& \frac{6}{16}& \frac{2}{16}& \frac{8}{16}\\ x^{2}P\left(X\right)& 0& \frac{6}{16}& \frac{4}{16}& \frac{10}{16}\end{array}$

Mean, $E\left(x\right)=\sum XP\left(X\right)=\frac{8}{16}$ or $\frac{1}{2},$

Var $\left(X\right)=\sum X^{2}P\left(X\right)-(\text{Mean}{)}^2=\frac{10}{16}-\frac{1}{4}=\frac{6}{16}$ or $\frac{3}{8}.$