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A pair of fair dice is rolled. What is the probability that the second dice lands on a higher value than does the first ?
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The correct answer is:
$\frac{5}{12}$
Possibilities of having higher number on second dice. $\begin{array}{ll}\text { First dice } & \text { Second dice }\end{array}$
$1 \quad 2,3,4,5,6 \rightarrow 5$ possibilities
$2 \quad 3,4,5,6 \rightarrow 4$ possibilities
$3 \quad 4,5,6 \rightarrow 3$ possibilities
$4 \quad 5,6 \rightarrow 2$ possibilities $5 \quad 6 \rightarrow 1$ possibility Total number of possibilities $=15$ Total number of events $=36$
$\therefore$ Probability $=\frac{15}{36}=\frac{5}{12}$
$1 \quad 2,3,4,5,6 \rightarrow 5$ possibilities
$2 \quad 3,4,5,6 \rightarrow 4$ possibilities
$3 \quad 4,5,6 \rightarrow 3$ possibilities
$4 \quad 5,6 \rightarrow 2$ possibilities $5 \quad 6 \rightarrow 1$ possibility Total number of possibilities $=15$ Total number of events $=36$
$\therefore$ Probability $=\frac{15}{36}=\frac{5}{12}$
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