Question

Two dice, one blue and one grey, are thrown at the same time.

(i) Write down all the possible outcomes and complete the following table:

Event : ‘Sum on 2 dice’	2	3	4	5	6	7	8	9	10	11	12
Probability	`1/36`						`5/36`				`1/36`

Two dice, one blue and one grey, are thrown at the same time. (i) Complete the following table:

‘Sum on 2 dice’	2	3	4	5	6	7	8	9	10	11	12
Probability	`1/36`						`5/36`				`1/36`

 (ii) A student argues that there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability `1/11` Do you agree with this argument? Justify your answer.

Answer 1

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 favourable 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`

‘Sum on 2 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 a different Probability for different outcome.