Find the probability of getting: sum 7

Two different dice are thrown simultaneously. Find the probability of getting:
(i) sum 7
(ii) sum ≤ 3
(iii) sum ≤ 10

Solution:

(i) Numbers whose sum is 7 will be (1, 6), (2, 5), (4, 3), (5, 2), (6, 1), (3, 4) = 6
P(E) = $\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$ = $\\ \frac { 6 }{ 36 }$ = $\\ \frac { 1 }{ 6 }$
(ii) Sum ≤ 3
Then numbers can be (1, 2), (2, 1), (1, 1) which are 3 in numbers
∴Probability will be
P(E) = $\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$ = $\\ \frac { 3 }{ 36 }$ = $\\ \frac { 1 }{ 12 }$
(iii) Sum ≤ 10
The numbers can be,
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, .6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5),
(6, 1), (6, 2), (6, 3), (6, 4) = 33
Probability will be
P(E) = $\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$ = $\\ \frac { 33 }{ 36 }$ = $\\ \frac { 11 }{ 12 }$