Two dice are thrown simultaneously. Find the probability of getting sum more than $7$

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Solution

Total outcome
$\begin{matrix} (1, 1) (1, 2) (1, 3) . . . . (2, 1) (2, 2) . . . (3, 1) (3, 2) . . . . . (4, 1) (a, 2) . . . . (6, 6) = 6 \times 6 = 36 \end{matrix}$
Favourable outcomes
$\begin{matrix} (2, 6) (3, 5) (3, 6) (4, 4) (4, 5) (4, 6) (5, 3) (5, 4) (5, 5,) (5, 6) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) = 15 \end{matrix}$
Probability
$\frac{15}{36} = \frac{5}{12}$

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