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(D) When two dice are rolled,the total number of possible outcomes is $6 	imes 6 = 36$.Let $X$ be the number on the first die and $Y$ be the number o

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$5/12$

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(D) When two dice are rolled,the total number of possible outcomes is $6 	imes 6 = 36$.Let $X$ be the number on the first die and $Y$ be the number on the second die.We want to find the probability $P(X The possible pairs $(X, Y)$ such that $X If $X=1$,$Y \in \{2, 3, 4, 5, 6\}$ ($5$ outcomes).If $X=2$,$Y \in \{3, 4, 5, 6\}$ ($4$ outcomes).If $X=3$,$Y \in \{4, 5, 6\}$ ($3$ outcomes).If $X=4$,$Y \in \{5, 6\}$ ($2$ outcomes).If $X=5$,$Y \in \{6\}$ ($1$ outcome).If $X=6$,there are no possible values for $Y$ such that $X Total favorable outcomes $= 5 + 4 + 3 + 2 + 1 = 15$.The required probability is $\frac{15}{36} = \frac{5}{12}$.

Two dice are rolled one after the other. The probability that the number on the first die is smaller than the number on the second die is

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