Refer to Example 13. (i) Complete the following table:
(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.
📚 Submitted Answer
Rahul (2025-06-30 10:03:56)
Answer:
(i) Complete the following table:
| 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 |
Explanation:
- Total possible outcomes when two dice are thrown = 6 × 6 = 36
- The number of ways to get each sum is different:
- 2 → (1,1)
- 3 → (1,2), (2,1)
- 4 → (1,3), (2,2), (3,1)
- 5 → (1,4), (2,3), (3,2), (4,1)
- 6 → (1,5), (2,4), (3,3), (4,2), (5,1)
- 7 → (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
- 8 → (2,6), (3,5), (4,4), (5,3), (6,2)
- 9 → (3,6), (4,5), (5,4), (6,3)
- 10 → (4,6), (5,5), (6,4)
- 11 → (5,6), (6,5)
- 12 → (6,6)
(ii) A student argues:
“There are 11 possible outcomes: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
So each has equal probability = 1/11.”
No, this argument is incorrect.
✔ Verified Answer
Share on WhatsApp