Complete the following table :

Event : 'Sum on two dice'	Probability
2	1/36
3
4
5
6
7
8	5/36
9
10
11
12	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 $\dfrac{1}{11}$ . Do you agree with this argument? Justify your answer.

On tossing two dice.

Sample space = {(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)}.

No. of possible outcomes = 36

Favourable outcomes for getting sum of 3 are {(1, 2), (2, 1)}

No. of favourable outcomes = 2

P(getting a sum of 3) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{36}$.

Favourable outcomes for getting sum of 4 are {(1, 3), (2, 2), (3, 1)}

No. of favourable outcomes = 3

P(getting a sum of 4) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{36}$.

Favourable outcomes for getting sum of 5 are {(1, 4), (2, 3), (3, 2), (4, 1)}

No. of favourable outcomes = 4

P(getting a sum of 5) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{4}{36}$.

Favourable outcomes for getting sum of 6 are {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}

No. of favourable outcomes = 5

P(getting a sum of 6) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{5}{36}$.

Favourable outcomes for getting sum of 7 are {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

No. of favourable outcomes = 6

P(getting a sum of 7) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{6}{36}$.

Favourable outcomes for getting sum of 9 are {(3, 6), (4, 5), (5, 4), (6, 3)}

No. of favourable outcomes = 4

P(getting a sum of 9) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{4}{36}$.

Favourable outcomes for getting sum of 10 are {(4, 6), (5, 5), (6, 4)}

No. of favourable outcomes = 3

P(getting a sum of 10) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{36}$.

Favourable outcomes for getting sum of 11 are {(5, 6), (6, 5)}

No. of favourable outcomes = 2

P(getting a sum of 11) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{36}$.

(ii) No, I don't agree because the argument is false. The eleven sums are not equally likely. Hence, each of them cannot have a probability of $\dfrac{1}{11}$. Also, in the above table we see that each sum has different probability.

Hence, the above argument is false.