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Q.
The probability of getting sum more than 7 when a pair of dice are thrown is:
Probability
Solution:
Here n (S) = $6^2$ = 36
Let E be the event "getting sum more than 7" i.e. sum of pair of dice = 8, 9, 10, 11, 12
i.e. $E = \begin{Bmatrix}\left(2,6\right)&\left(3,5\right)&\left(4,4\right)&\left(5,3\right)&\left(6,2\right)\\ \left(3,6\right)&\left(4,5\right)&\left(5,4\right)&\left(6,3\right)&\\ \left(4,6\right)&\left(5,5\right)&\left(6,4\right)&&\\ \left(5,6\right)&\left(6,5\right)&\left(6,6\right)&&\end{Bmatrix}$
$\therefore \ n(E) = 15$
$\therefore $ Required prob $ = \frac{n(E)}{n(S)} = \frac{15}{36} = \frac{5}{12}$