Two unbiased dice are thrown. Then the probab | vedclass

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(D) The total number of outcomes when two dice are thrown is $6 	imes 6 = 36$. Let $A$ be the event of getting a doublet. The outcomes are $(1,1), (2

Vedclass · Accepted Answer

$\frac{7}{9}$

Vedclass · Answer

(D) The total number of outcomes when two dice are thrown is $6 	imes 6 = 36$. Let $A$ be the event of getting a doublet. The outcomes are $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$,so $n(A) = 6$. Let $B$ be the event of getting a total of $10$. The outcomes are $(4,6), (5,5), (6,4)$,so $n(B) = 3$. The intersection $A \cap B$ is the outcome $(5,5)$,so $n(A \cap B) = 1$. The number of outcomes that are either a doublet or a total of $10$ is $n(A \cup B) = n(A) + n(B) - n(A \cap B) = 6 + 3 - 1 = 8$. The number of outcomes that are neither a doublet nor a total of $10$ is $36 - 8 = 28$. Therefore,the required probability is $\frac{28}{36} = \frac{7}{9}$.

Two unbiased dice are thrown. Then the probability that neither a doublet nor a total of $10$ will appear is

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