Refer to Example 13. (i) 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 (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.
Complete the following table: Event: ‘Sum on 2 dice’ | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12. Probability | 1/36 | ...
| Event: '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, I do not agree with this argument. The 11 outcomes (sums 2, 3, 4, ..., 12) are not equally likely. When two dice are thrown, there are 36 possible outcomes ( = 36). Different sums have different numbers of favourable outcomes. For example, sum = 7 can occur in 6 ways: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), so P(7) = 6/36. Whereas sum = 2 occurs in only 1 way: (1,1), so P(2) = 1/36. Since the outcomes are not equally likely, the probability is not 1/11 for each sum.
Explanation
For part (i), students need to count favourable outcomes for each sum from the 36 total outcomes. The pattern is symmetric: 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 favourable outcomes for sums 2 through 12 respectively. For part (ii), the key concept is that elementary events must be equally likely for equal probability distribution. The sums are compound events with varying numbers of favourable outcomes.