If three dice are thrown, the probability that they show the numbers in A.P. is

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If three dice are thrown, the probability that they show the numbers in A.P. is

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$\frac{5}{18}$

$\frac{1}{36}$

$\frac{1}{12}$

$\frac{2}{9}$

answer is B.

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Detailed Solution

$n (S) = 6^{3} = 216$

E= numbers in A.P

$A.P. with common difference one are = 4 = (123), (234), (345), (456)$

$A.P. with common difference two are = 2 = (135), (246)$

$A.P. with common difference zero are = 6$

$n (E) = 2 + 4 + 6 = 12$

$Required probability P = \frac{12}{216} = \frac{1}{18}$

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