Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is :

(i) 0

(ii) 12

(iii) less than 12

(iv) less than or equal to 12

(i) Total number of possible outcomes = 36 (i.e., (1, 1), (1, 2), (1, 3), (1, 4), ……………..(6, 4), (6, 5), (6, 6))

Number of favourable outcomes (where the sum of the two numbers appearing on the top of the dice is 0) = 0

P(where the sum of the two numbers appearing on the top of the dice is 0) = $\dfrac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

= $\dfrac{0}{36}$

= 0

Hence, the probability of the sum of the two numbers on the top of the dice being 0 is 0.

(ii) Total number of possible outcomes = 36 (i.e., (1, 1), (1, 2), (1, 3), (1, 4), ……………..(6, 4), (6, 5), (6, 6))

Number of favourable outcomes (where the sum of the two numbers appearing on the top of the dice is 12) = 1 ((6, 6))

P(where the sum of the two numbers appearing on the top of the dice is 12) = $\dfrac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

= $\dfrac{1}{36}$

Hence, the probability of the sum of the two numbers on the top of the dice being 12 is $\dfrac{1}{36}$.

(iii) Total number of possible outcomes = 36 (i.e., (1, 1), (1, 2), (1, 3), (1, 4), ……………..(6, 4), (6, 5), (6, 6))

Number of favourable outcomes (where the sum of the two numbers appearing on the top of the dice is less than 12) = 35 ((1, 1), (1, 2), (1, 3), …………..(6, 5))

P(where the sum of the two numbers appearing on the top of the dice is less than 12) = $\dfrac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

= $\dfrac{35}{36}$

Hence, the probability of the sum of the two numbers on the top of the dice being less than 12 is $\dfrac{35}{36}$.

(iv) Total number of possible outcomes = 36 (i.e., (1, 1), (1, 2), (1, 3), (1, 4), ……………..(6, 4), (6, 5), (6, 6))

Number of favourable outcomes (where the sum of the two numbers appearing on the top of the dice is less than or equal to 12) = 36 ((1, 1), (1, 2), (1, 3), …………..(6, 5), (6, 6))

P(where the sum of the two numbers appearing on the top of the dice is less than or equal to 12) = $\dfrac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

= $\dfrac{36}{36}$

= 1

Hence, the probability of the sum of the two numbers on the top of the dice being less than or equal to 12 is 1.