Question

Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the probability distribution of the number of kings.

Answer 1

$P(X=x)={}^n{C}_x{q}^{n-x}{p}^x$

Here, $n=$ number of cards drawn $=2$

$p=$ Probability of getting king $=\frac{4}{52}=\frac{1}{13}$

$q=1-p=1-\frac{1}{13}=\frac{12}{13}$

Hence,

$\Rightarrow$ $P(X=x)={}^2{C}_x{\left(\frac{1}{13}\right)}^x{\left(\frac{12}{13}\right)}^{2-x}$ ----- ( 1 )

Since $2$ cards are drawn,

We can get $0$ ace, $1$ king or $2$ king.

So, $X$ can be $0,1,2$

Putting values in ( 1 )

$P(X=0)={}^2{C}_0{\left(\frac{1}{13}\right)}^0{\left(\frac{12}{13}\right)}^{2-0}=1\times 1\times {\left(\frac{12}{13}\right)}^2=\frac{144}{169}$

$P(X=1)={}^2{C}_1{\left(\frac{1}{13}\right)}^1{\left(\frac{12}{13}\right)}^{2-1}=2\times \frac{1}{13}\times \frac{12}{13}=\frac{24}{169}$

$P(X=2)={}^2{C}_2{\left(\frac{1}{13}\right)}^2{\left(\frac{12}{13}\right)}^{2-2}=1\times {\left(\frac{1}{13}\right)}^2=\frac{1}{169}$

$X$	$0$	$1$	$2$
$P\left(X\right)$	$\frac{144}{169}$	$\frac{24}{169}$	$\frac{1}{169}$