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.