Two cards are drawn successively with replacement from a well-shuffle…

The correct answer is Option (1) → 

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

Probability of drawing an ace from a pack of 52 cards

$=p=\frac{4}{52}=\frac{1}{13}$, so $q=1-\frac{1}{13}=\frac{12}{13}$.

As the cards are drawn successively with replacement, events are independent, therefore, it is a problem of binomial distribution with

$p =\frac{1}{13},q = \frac{12}{13}$ and $n = 2$.

If X denotes the number of aces drawn in a draw of 2 cards, then X can take values 0, 1, 2.

$P(0) = {^2C}_0 q^2 = 1.(\frac{12}{13})^2=\frac{144}{169}$,

$P(1) = {^2C}_1 pq = 2.\frac{12}{13}.\frac{12}{13}=\frac{24}{169}$ and

$P(2) = {^2C}_2 p^2 = 1.(\frac{1}{13})^2=\frac{1}{169}$

∴ Required probability distribution is $\begin{pmatrix}0&1&2\\\frac{144}{169}&\frac{24}{169}&\frac{1}{169}\end{pmatrix}$.