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 | \(\frac{1}{36}\) | \(\frac{5}{36}\) | \(\frac{1}{36}\) |
(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.
(i) Probabilities of sums 2 to 12 respectively are: \(\dfrac{1}{36}\), \(\dfrac{2}{36}\), \(\dfrac{3}{36}\), \(\dfrac{4}{36}\), \(\dfrac{5}{36}\), \(\dfrac{6}{36}\), \(\dfrac{5}{36}\), \(\dfrac{4}{36}\), \(\dfrac{3}{36}\), \(\dfrac{2}{36}\), \(\dfrac{1}{36}\).
(ii) No. The sums are not equally likely; for example, 7 can occur in 6 different ways, while 2 or 12 can occur in only 1 way each, so their probabilities are not all 1/11.