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Questions
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.
Solution
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.
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