Question 5
A fair die with faces 1, 2, 3, 4, 5 and 6 is tossed twice. Calculate the probability that the sum of numbers that showed up is:
- a multiple of 3.
- between 3 and 6.
Observation
The Chief Examiner reported that majority of the candidates attempted this question satisfactorily.
In part (a), they were able to get the table showing the possible outcomes
⊕ | 1 | 2 | 3 | 4 | 5 | 6 |
1 | 2 | 3 | 4 | 5 | 6 | 7 |
2 | 3 | 4 | 5 | 6 | 7 | 8 |
3 | 4 | 5 | 6 | 7 | 8 | 9 |
4 | 5 | 6 | 7 | 8 | 9 | 10 |
5 | 6 | 7 | 8 | 9 | 10 | 11 |
6 | 7 | 8 | 9 | 10 | 11 | 12 |
P( a multiple of 3) =12⁄36 = 1⁄3.
In part (b), they also obtained P(between 3 and 6) = 7⁄36.