Consider A and B be the events when the sum is obtained as 7 and 11 respectively. Number of times sum 7 and 11 occurs, n(A)={(2,5),(5,2),(3,4),(4,3),(1,1),(6,1)}=6 n(B)={(5,6),(6,5)}=2 Now, the probability is, P(A)=‌
6
36
=‌
1
6
P(B)=‌
2
36
=‌
1
18
The probability when neither the sum of 7 or 11 occurs, P(C)=‌
36−(6+2)
36
=‌
7
9
The probability that 7 comes before 11 P‌‌=‌