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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.

Answers (1)

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

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

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

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

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

Posted by

Ravindra Pindel

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