Question

Two cards are drawn simutaneously (or successively without replacement) from a well shuffled pack of $52$ cards. Then standard deviation of the number of kings:

• A. $0.37$
• B. $0.27$
• C. $0.47$
• D. None of these

Answer 1

The correct option is A $0.37$

Let $X$ denotes the numbr of kings in a draw of two cards. $X$ is a random variable which can assume the values $0,1$ or $2$
Now, $P(X=0)=P\left(\text{no king}\right)=\frac{{}^{48}{C}_2}{{}^{52}{C}_2}=\frac{\frac{48!}{2!(48-2)!}}{\frac{52!}{2!(52-2)!}}=\frac{48}{52}\times \frac{47}{51}=\frac{188}{221}$
$P(X=1)=P\left(\text{no king and one king}\right)=\frac{{}^4{C}_1{}^{48}{C}_1}{{}^{52}{C}_2}=\frac{4\times 48\times 2}{52\times 51}=\frac{32}{221}$
and $P(X=2)=P\left(\text{two king}\right)=\frac{{}^4{C}_2}{{}^{52}{C}_2}=\frac{4\times 3}{52\times 51}=\frac{1}{221}$
Thus, the probability distribution of $X$ is
$\begin{array}{cccc}\text{X}& \text{0}& \text{1}& \text{2}\\ \text{P(X)}& \frac{188}{221}& \frac{32}{221}& \frac{1}{221}\end{array}$
Mean of $X=E\left(X\right)=\sum _{i=1}^n{x}_{i}p\left(x_i\right)$
$=0\times \frac{188}{221}+1\times \frac{32}{221}+2\times \frac{1}{221}=\frac{34}{221}$
$E\left(X^2\right)=\sum _{i=1}^n{x}_i^{2}p\left(x_i\right)$
$=0^2\times \frac{188}{221}+1^2\times \frac{32}{221}+2^2\times \frac{1}{221}=\frac{36}{221}$
$Var\left(X\right)=E\left(X^2\right)-\left(E\right(X){)}^2=\frac{36}{221}-{\left(\frac{34}{221}\right)}^2=\frac{6800}{(221{)}^2}$
Therefore ${\sigma }_x=\sqrt{Var\left(X\right)}=\frac{\sqrt{6800}}{221}=0.37$