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

Let S = no. of Kings

and F = number of non-kings

$\therefore P(S) = \frac{4}{52}$ and $P(F) = \frac{48}{52}$

Let $x_i$ be the event that the king is drawn

 $x_i$ $Px_i$ $x_iPx_i$ $x_i^2$ $Px_i \cdot x_i^2$ 0 $\frac{48}{52}\times \frac{47}{51}\times 1 = \frac{2256}{2652}$ 0 0 0 1 $\frac{4}{52}\times \frac{48}{51}\times 2 = \frac{384}{2652}$ $\frac{384}{2652}$ 1 $\frac{384}{2652}$ 2 $\frac{4}{52}\times \frac{3}{51}\times 1 = \frac{12}{2652}$ $\frac{12}{2652}$ 4 $\frac{48}{2652}$

 $Px_i$ $\frac{2256}{2652}$ $\frac{384}{2652}$ $\frac{12}{2652}$ $x_i$ 0 1 2

Mean    $\overline{x} = \sum x_i^2P(x_i)$

$\overline{x} =\frac{384 + 12}{2652} = \frac{396}{2652}$

Variance $\sigma^2= \sum x_i^2P(x_i)-(\overline{x})^2$

$= \frac{432}{2652} -\frac{396}{2652}^2$

$= \frac{36}{221} -\frac{33}{221}^2$

$\Rightarrow \frac{(36\times 221)-(33\times 33)}{221\times 221}$

$\Rightarrow \frac{7956 - 1089}{48841} = \underline{\frac{6867}{48841}}$

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