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Q.

Let N be the sum of the numbers appeared when two fair dice are rolled and let the probability that $N - 2, \sqrt{3 N}, N + 2$ are in a geometric progression be $\frac{k}{48}$ Then the value of $k$ is

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a

16

b

2

c

4

d

8

answer is D.

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Detailed Solution

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Given $N - 2, \sqrt{3} N, N + 2$ are in G.P then $(\sqrt{3} N)^{2} = (N - 2) (N + 2) \Rightarrow 3 N = N^{2} - 4$

$\Rightarrow N^{2} - 3 N - 4 = 0 \to (N - 4) (N + 1) = 0$

$N = 4 \to Event of getting sum 4 = \{(1, 3), (3, 1), (2, 2)\}$

$then P (X = 4) = \frac{3}{36} = \frac{1}{12} = \frac{4}{48}$
$then = k = 4$

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