Two cards are drawn simultaneously (or successively without replacement) from a well shuffled pack of $52$ cards. Find the mean, variance and standard deviation of the number of kings.

Hint: In order to solve these types of sums, the student has to have thorough knowledge about Permutation and combinations. Apart from this the student should also be well-versed with what a pack of cards consists of. He should know what a face card is, which are those face cards from a pack of $52$ cards. This is because solving a sum on probability is fully based on common sense and the student just needs to know about the possibilities. In this sum the student needs to list down the possibilities of obtaining a King in $2$ drawings. Then based on this date he has to calculate mean variance and standard deviation. Also for this particular sum the student needs to know the formula for the mean, variance and standard deviation when it is related to Probability.

Complete answer:
Let $x$ denote the number of King in a draw of two cards.
$\therefore x = 0$, means no King card .
$\Rightarrow P(x = 0) = \dfrac{{^{48}{C_2}}}{{^{52}{C_2}}}$,
This is because we have removed $4$king cards as we are assuming that there will be no king drawn.
$\Rightarrow P(x = 0) = \dfrac{{48 \times 47}}{{52 \times 51}}$=$\dfrac{{188}}{{221}}.............(1)$
Now finding \[P(x = 1)\], probability of obtaining only one king card out of four cards, and the remaining card is from the $48$cards from the pack.
\[\Rightarrow P(x = 1) = \dfrac{{^{48}{C_1}{ \times ^4}{C_1}}}{{^{52}{C_2}}}\]
\[\Rightarrow P(x = 1) = \dfrac{{4 \times 48}}{{52 \times 51}}\]=$\dfrac{{32}}{{221}}.............(2)$
Now finding \[P(x = 2)\], probability of obtaining $2$ king card out of four cards,
$\Rightarrow P(x = 2) = \dfrac{{^4{C_2}}}{{^{52}{C_2}}}$
Simplifying the above equation
$\Rightarrow P(x = 2) = \dfrac{{^4{C_2}}}{{^{52}{C_2}}} = \dfrac{1}{{221}}..........(3)$
Now using equation $1,2\& 3$ we will find mean, variance and standard deviation.
Mean for the probability is given by $\sum\limits_{x = 2}^n {x \times P(x)} $
$\Rightarrow Mean(E(x)) = 0 \times \dfrac{{18}}{{221}} + 1 \times \dfrac{{32}}{{221}} + 2 \times \dfrac{1}{{221}}$
$\Rightarrow Mean = \dfrac{{34}}{{221}}.........(4)$
Variance for the probability is given by $E({x^2}) - {(E(x))^2}$
$E({x^2}) = \sum\limits_{x = 1}^n {{x^2} \times P(x)} $ = ${0^2} \times \dfrac{{18}}{{221}} + {1^2} \times \dfrac{{32}}{{221}} + {2^2} \times \dfrac{1}{{221}}$
$\Rightarrow E({x^2}) = \dfrac{{36}}{{221}}$
Variance =$\dfrac{{36}}{{221}} - {(\dfrac{{34}}{{221}})^2}$=$\dfrac{{6800}}{{{{221}^2}}}............(5)$
Standard deviation for the probability is given by $\sqrt {Variance} $
$\therefore \text{Standard Deviation} = \sqrt {\dfrac{{6800}}{{{{221}^2}}}} $=$0.37......(6)$

Thus from equation $4,5,6$ we get the desired answer.

Note: Only thing important in this sum was the application of probability to find Mean variance and standard Deviation. Students should remember this formula while calculating mean ,variance & standard deviation for probability sums. While solving sums related to probability students should use permutation and combination properly otherwise the entire sum would go wrong.