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