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

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$1 / 2$

$7 / 18$

$3 / 4$

$5 / 12$

answer is D.

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Detailed Solution

Consider the following events

$A_{i} \to$ Getting number $i$ on first die

$B_{i} \to$ Getting a number more than $i$ on second die

Required probability

$\begin{array}{l} = P (A_{1} \cap B_{1}) + P (A_{2} \cap B_{2}) + P (A_{3} \cap B_{3}) \\ + P (A_{4} \cap B_{4}) + P (A_{5} \cap B_{5}) \\ = \sum_{i = 1}^{5} P (A_{i} \cap B_{1}) \end{array}$

$= \sum_{i = 1}^{5} P (A_{i}) P (B_{i}) [∵ A_{i}, B_{i} are independent]$

$\begin{aligned} = \frac{1}{6} [P (B_{1}) + P (B_{2}) + \dots + P (B_{5})] \\ = \frac{1}{6} (\frac{5}{6} + \frac{4}{6} + \frac{3}{6} + \frac{2}{6} + \frac{1}{6}) = \frac{5}{12} \end{aligned}$

$A L I T E R$ There are 36 elementary events associated with the random experiment of throwing a pair of dice. Out of these 36 elementary events there are six elementary events in which numbers on two dice are equal and in the remaining 30 we have 15 elementary events in which number on first die is greater than the number on second die and in other 15 the number on first die is less than that on second.

Hence, required probability $= \frac{15}{36} = \frac{5}{12}$

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