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: [Table with probabilities for sums 2-12]
| 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. When two dice are thrown, there are 36 possible outcomes ( = 36), not 11. The 11 outcomes mentioned (sums 2 to 12) are not equally likely. For example, sum 7 can occur in 6 ways: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), while sum 2 can occur in only 1 way: (1,1). Since the outcomes are not equally likely, the probability of each sum is not 1/11. The correct probability is calculated as P(E) = Number of favourable outcomes / Total number of possible outcomes.
Explanation
This question tests understanding of probability when two dice are thrown. The total number of possible outcomes is 36 (). For part (i), students must count favourable outcomes for each sum. For part (ii), the key concept is that the 11 sums are NOT equally likely outcomes - the elementary outcomes are the 36 ordered pairs. The student's argument incorrectly assumes all sums have equal probability.