In a simultaneous throw of a pair of dice, find the probability of getting a total more than 7. · ABC26GN5332

Here, $n(S)=(6 \times 6)=36$. Let $E=$ Event of getting a total more than 7 $=\{(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)\}$. $\therefore P(E)=\frac{n(E)}{n(S)}=\frac{15}{36}=\frac{5}{12}$.