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 \( \frac{1}{11} \). Do you agree with this argument? Justify your Solution:.

If 2 dices are thrown, the possible events are:

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

So, the total numbers of events: 6 × 6 = 36

(i) It is given that to get the sum as 2, the probability is \( \frac{1}{36} \) as the only possible outcomes = (1,1)

For getting the sum as 3, the possible events (or outcomes) = E (sum 3) = (1,2) and (2,1)

So, P(sum 3) \( = \ \frac{2}{36} \)

Similarly,

E (sum 4) = (1,3), (3,1), and (2,2)

So, P (sum 4) \( = \ \frac{3}{36} \)

E (sum 5) = (1,4), (4,1), (2,3), and (3,2)

So, P (sum 5) \(= \ \frac{4}{36} \)

E (sum 6) = (1,5), (5,1), (2,4), (4,2), and (3,3)

So, P (sum 6) \( = \ \frac{5}{36} \)

E (sum 7) = (1,6), (6,1), (5,2), (2,5), (4,3), and (3,4)

So, P (sum 7) \(= \ \frac{6}{36} \)

E (sum 8) = (2,6), (6,2), (3,5), (5,3), and (4,4)

So, P (sum 8) \(= \ \frac{5}{36} \)

E (sum 9) = (3,6), (6,3), (4,5), and (5,4)

So, P (sum 9) \(= \ \frac{4}{36} \)

E (sum 10) = (4,6), (6,4), and (5,5)

So, P (sum 10) \( = \ \frac{3}{36} \)

E (sum 11) = (5,6), and (6,5)

So, P (sum 11) \( = \ \frac{2}{36} \)

E (sum 12) = (6,6)

So, P (sum 12) \( = \ \frac{1}{36} \)

So, the table will be as:

(ii) The argument is not correct as it is already justified in (i) that the number of all possible outcomes is 36 and not 11.