Two cards are drawn simultaneously from a well shuffled pack of 52 cards. Then variance of the number of kings is

$\frac{6800}{(221)^2}$

let X → denote number of kings in a draw of 2 cards.

for no king $P (X = 0) =\frac{{ }^{48} C_2}{{ }^{52} C_2}=\frac{48 ! 2 ! 50 !}{2 ! 46 ! 52}=\frac{48 \times 47}{52 \times 51}=\frac{188}{221}$

for 1 king $P(X = 1) = \frac{4 C_1 \times 48 C_1}{{ }^{52} C_2}=\frac{2\times 4 \times 48}{52 \times 51}=\frac{32}{221}$

for 2 teings $P(x=2)=\frac{{ }^4 C_2}{{ }^{52} C_2}=\frac{4 \times 3}{52 \times 51}=\frac{1}{221}$

Probability distribution of kings

Mean → E(X) = $\sum p\left(x_i\right) \times x_i=\frac{0 \times 188}{221}+\frac{1 \times 32}{221}+\frac{2 \times 1}{221}$

$ =\frac{34}{221}$

$E\left(x^2\right)=\sum p\left(x_i\right) x_i^2 = 0^2 \times \frac{188}{221}+1^2 \times \frac{32}{221}+2^2 \times \frac{1}{221}=\frac{36}{221}$

Variance → $\sigma^2=E\left(x^2\right)-(E(x))^2=\frac{36}{221}-\left(\frac{34}{221}\right)^2=\frac{6800}{(221)^2}$