Question

Two cards are drawn simultaneously (or successively without replacement) from a well shuffled pack of $52$ cards. Find mean variance and standard deviation of the number of kings.

Answer 1

Let X denote the number of King in a draw of two card

$\begin{array}{c}\Rightarrow P\left(X=0\right)=P\left(noking\right)=\frac{48c_2}{52c_2}=\frac{48\times 47}{52\times 51}=\frac{188}{221}\\ \Rightarrow P\left(X=1\right)=P\left(onekingandonenonking\right)=\frac{4c_1\times 48c_1}{52c_2}=\frac{4\times 48}{52\times 51}=\frac{32}{221}\\ \Rightarrow P\left(X=2\right)=P\left(twoking\right)=\frac{4c_2}{52c_2}=\frac{4\times 3}{52\times 51}=\frac{1}{221}\end{array}$

Probability of distribution

$\begin{array}{c}\Rightarrow MeanofX={\sum }_{2=1}^{n}xip\left(xi\right)=0\times \frac{188}{221}+1\times \frac{32}{221}+\frac{2}{221}=\frac{34}{221}\\ \Rightarrow E\left(X^2\right)={\sum }_{i=1}^{n}x{i}^{2}p\left(xi\right)=0^2\times \frac{188}{221}+1^2\times \frac{32}{221}+2^2\times \frac{1}{221}=\frac{36}{221}\\ \Rightarrow Var\left(X\right)=E\left(X^2\right)-{\left[E\left(X\right)\right]}^2=\frac{36}{221}-{\left(\frac{34}{221}\right)}^2=\frac{6800}{{\left(221\right)}^2}\\ 6\times \sqrt{Var\left(X\right)}=\frac{\sqrt{6800}}{221}=0.37\end{array}$