Two dice are thrown at the same time. The probability of getting a sum of 11 is:

The probability of an event E occurring is the ratio of the number of favourable cases to the total number of cases.

$P\left(E\right)=\frac{\text{ no. of cases favourable to event }E}{\text{ Total no. of cases }}$

When 2 dice are thrown simultaneously a sum of 15 can be obtained by the following combinations:

(1,1) (2,1) (3,1) (4,1) (5,1) (6,1)

​(1,2) (2,2) (3,2) (4,2) (5,2) (6,2)

​(1,3) (2,3) (3,3) (4,3) (5,3) (6,3)

​(1,4) (2,4) (3,4) (4,4) (5,4) (6,4)

​(1,5) (2,5) (3,5) (4,5) (5,5) (6,5)

​(1,6) (2,6) (3,6) (4,6) (5,6) (6,6)​

Favourable outcomes = (5,6), and (6,5)

Therefore the probability is:

$=\frac{2}{36}$

$=\frac{1}{18}$