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Q.

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

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a

p + q = 1

b

p < q

c

p = q

d

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

answer is B, D.

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Detailed Solution

$\begin{array}{l} p = \frac{4 C_{2} \times 2 (49)!}{52!} = \frac{12}{52 \times 51 \times 50} \\ q = \frac{49 c_{4} \times 4 \times 48!}{52!} \\ = \frac{49 \times 48 \times 47 \times 46}{52 \times 51 \times 50 \times 49} = \frac{48 \times 47 \times 46}{52 \times 51 \times 50} \end{array}$

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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