Three cards are drawn successively with replacement from a well shuff…

The correct answer is Option (1) → 

Probability Distribution:

$X$	0	1	2	3
$P(X)$	$\frac{27}{64}$	$\frac{27}{64}$	$\frac{9}{64}$	$\frac{1}{64}$

Mean: $\frac{3}{4}$

Probability of drawing a spade from a pack of 52 cards = $p =\frac{13}{52}=\frac{1}{4}$

so, $q = 1-p=1-\frac{1}{4}=\frac{3}{4}$

As the cards are drawn successively with replacement, events are independent, therefore, it is a problem of binomial distribution with $p = \frac{1}{4},q=\frac{3}{4}$ and $n = 3$.

Let random variable X denote the number of spades, so X can take values 0, 1, 2, 3.

$P(X = 0) = {^3C}_0q^3=1.(\frac{3}{4})^3=\frac{27}{64}$

$P(X = 1) = {^3C}_1 p.q^2=3.\frac{1}{4}(\frac{3}{4})^2=\frac{27}{64}$

$P(X = 2) = {^3C}_2 p^2q^1=3(\frac{1}{4})^2.\frac{3}{4}=\frac{9}{64}$

$P(X = 3) = {^3C}_3 p^3 =1.(\frac{1}{4})^3=\frac{1}{64}$

∴ Required probability distribution is $\begin{pmatrix}0&1&2&3\\\frac{27}{64}&\frac{27}{64}&\frac{9}{64}&\frac{1}{64}\end{pmatrix}$

Mean = $Σp_ix_i=\frac{1}{64}(27 × 0+27 × 1+9×2+1×3)$

$=\frac{1}{64}(0+27+18+3)=\frac{48}{64}=\frac{3}{4}$