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

If 2 dice 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 number of events is 6 × 6 = 36

(i) From the table, the probability of the sum as 2, is 1/36 as (1,1) only possible outcomes

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

Hence, P(sum 3) = 2/36

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

Hence, P (sum 4) = 3/36

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

Hence, P (sum 5) = 4/36

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

Hence, P (sum 6) = 5/36

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

Hence, P (sum 7) = 6/36

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

Hence, P (sum 8) = 5/36

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

Hence, P (sum 9) = 4/36

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

Hence, P (sum 10) = 3/36

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

Hence, P (sum 11) = 2/36

Similarly, E (sum 12) = (6,6)

Hence, P (sum 12) = 1/36

(ii) The argument is incorrect because it is already supported in (i) which states that there are 36 possible outcomes rather than 11.