Two cards are drawn successively without replacement from a well shuffled deck of cards. Find the mean and standard variation of the random variable $X$, where $X$ is the number of aces.

Let $X$ denotes a random variable of number of aces.
$\therefore$ $X = 0.1.2$

Now, $\quad P(X = 0) = \frac{{48}}{{52}} \cdot \frac{{47}}{{51}} = \frac{{2256}}{{2652}}$

$P(X = 1) = \frac{{48}}{{52}} \cdot \frac{4}{{51}} + \frac{4}{{52}} \cdot \frac{{48}}{{51}} = \frac{{384}}{{2652}}$

$P(X = 2) = \frac{4}{{52}} \cdot \frac{3}{{51}} = \frac{{12}}{{2652}}$

We know that, Mean $(\mu ) = E(X) = \Sigma XP(X)$
$= 0 + \frac{{384}}{{2652}} + \frac{{24}}{{2652}}$
$= \frac{{408}}{{2652}} = \frac{2}{{13}}$

Also, ${\mathop{\rm Var}\nolimits} (X) = E\left( {{X^2}} \right) - {[E(X)]^2} = \Sigma {X^2}P(X) - {[E(X)]^2}$

$= \left[ {0 + \frac{{384}}{{2652}} + \frac{{48}}{{2652}}} \right] - {\left( {\frac{2}{{13}}} \right)^2}$

$= \frac{{432}}{{2652}} - \frac{4}{{169}} = 0.1628 - 0.0236 = 0.1391$

$\therefore$ $\quad$ Standard deviation $= \sqrt {{\mathop{\rm Var}\nolimits} (X)} = \sqrt {0.139} = 0.373$ (approx)