Total no. of possible outcomes when 2 dice are thrown = 6×6=36 which are
{ (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) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) }
E ⟶ event of getting sum on 2 dice as 2
No. of favourable outcomes = 1{(1, 1)}
Probability, P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes)
P(E) = 1/36
E ⟶ event of getting sum as 3
No. of favourable outcomes = 2 {(1, 2) (2, 1)}
P(E) = 2/36
E ⟶ event of getting sum as 4
No. of favourable outcomes = 3 {(3, 1) (2, 2) (1, 3)}
P(E) = 3/36
E ⟶ event of getting sum as 5
No. of favourable outcomes = 4 {(1, 4) (2, 3) (3, 2) (4, 1)}
P(E) = 4/36
E ⟶ event of getting sum as 6
No. of favourable outcomes = 5 {(1, 5) (2, 4) (3, 3) (4, 2) (5, 1)}
P(E) = 5/36
E ⟶ event of getting sum as 7
No. of favourable outcomes = 6 {(1, 6) (2, 5) (3, 4) (4, 3) (5, 2) (6, 1)}
P(E) = 6/36
E ⟶ event of getting sum as 8
No. of favourable outcomes = 5 {(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)}
P(E) = 5/36
E ⟶ event of getting sum as 9
No. of favourable outcomes = 4 {(3, 6) (4, 5) (5, 4) (6, 3)}
P(E) = 4/36
E ⟶ event of getting sum as 10
No. of favourable outcomes = 3 {(4, 6) (5, 5) (6, 4)}
P(E) = 3/36
E ⟶ event of getting sum as 11
No. of favourable outcomes = 2 {(5, 6) (6, 5)}
P(E) = 2/36
E ⟶ event of getting sum as 12
No. of favourable outcomes = 1 {(6, 6)}
P(E) = 1/36
Event ‘Sum
on two dice’ |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
Probability |
1/36 |
2/36 |
3/36 |
4/36 |
5/36 |
6/36 |
5/36 |
4/36 |
3/36 |
2/36 |
1/36 |
No, the outcomes are not equally likely from the above table we see that, there is different probability for different outcome.