Let p be the probability that in a pack of playing cards two kings are adjacent and q be the probability that no two kings are together, then

p = q

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p < q

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p + q = 1

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q = \frac{48 \times 47 \times 46}{52 \times 51 \times 50}

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Solution

The correct option is A $p < q$

$C a s e - I$ : At least two kings are together. Consider two kings together as a single card.
Number of such permutations = $(52 - 1)! = 51!$
$C a s e - I I$ : At least three kings are together. Consider three kings together as a single card.

Number of such permutation = $(52 - 2)! = 50!$
$P r o b a b i l i t y t h a t t w o k i n g s a r e a d j a c e n t = p$
$\Rightarrow p = \frac{51! - 50!}{52!} = \frac{51 - 1}{51 \times 52} = \frac{25}{1326} .$

$P r o b a b i l i t y t h a t n o t w o k i n g s a r e a d j a c e n t = q$
$\Rightarrow q = 1 - \frac{51!}{52!} = \frac{51}{52}$

$\Rightarrow p < q [B]$

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