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(D) When two dice are rolled,the total number of possible outcomes is $6 	imes 6 = 36$. Let the outcome of the first die be $x$ and the second die be

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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 the outcome of the first die be $x$ and the second die be $y$. We want the probability that $x The possible pairs $(x, y)$ such that $x For $x=1$,$y \in \{2, 3, 4, 5, 6\}$ ($5$ outcomes). For $x=2$,$y \in \{3, 4, 5, 6\}$ ($4$ outcomes). For $x=3$,$y \in \{4, 5, 6\}$ ($3$ outcomes). For $x=4$,$y \in \{5, 6\}$ ($2$ outcomes). For $x=5$,$y \in \{6\}$ ($1$ outcome). For $x=6$,there are no possible values for $y$. Total favorable outcomes $= 5 + 4 + 3 + 2 + 1 = 15$. The 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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