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(D) $1$. The total number of outcomes when two dice are thrown is $6 	imes 6 = 36$.$2$. $A$ doublet is $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$. Th

Vedclass · Accepted Answer

$671$

Vedclass · Answer

(D) $1$. The total number of outcomes when two dice are thrown is $6 	imes 6 = 36$.$2$. $A$ doublet is $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$. There are $6$ such outcomes.$3$. Probability of success $p = \frac{6}{36} = \frac{1}{6}$.$4$. Probability of failure $q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$.$5$. The number of trials $n = 4$. The probability of getting at least $1$ success is $P(X \ge 1) = 1 - P(X = 0)$.$6$. $P(X = 0) = ^nC_0 \cdot p^0 \cdot q^n = 1 \cdot 1 \cdot (\frac{5}{6})^4 = \frac{625}{1296}$.$7$. $P(X \ge 1) = 1 - \frac{625}{1296} = \frac{1296 - 625}{1296} = \frac{671}{1296}$.

Two dice are thrown successively $4$ times and getting a doublet is called a success. The probability of getting at least $1$ success is (in $/1296$)

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