Two dice (each bearing numbers 1 to 6) are rolled together. Find the probability that the sum of the numbers on the upper-most faces of two dice is :

(i) 4 or 5

(ii) 7, 8 or 9.

(iii) between 5 and 8.

(iv) more than 10.

(v) less than 6.

When two dice are rolled together;

No. of possible outcomes = 6 × 6 = 36.

(i) Favourable outcomes for sum of numbers on the upper-most faces of two dice to be 4 or 5 are :

(1, 3), (1, 4), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1).

∴ No. of favourable outcomes = 7.

P(getting a sum of 4 or 5) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{7}{36}$.

Hence, the probability of getting sum of numbers on the upper-most faces of two dice to be 4 or 5 = $\dfrac{7}{36}$.

(ii) Favourable outcomes for sum of numbers on the upper-most faces of two dice to be 7, 8 or 9 are :

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

∴ No. of favourable outcomes = 15.

P(getting a sum of 7, 8 or 9) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{15}{36} = \dfrac{5}{12}$.

Hence, the probability of getting sum of numbers on the upper-most faces of two dice to be 7, 8 or 9 = $\dfrac{5}{12}$.

(iii) Favourable outcomes for sum of numbers on the upper-most faces of two dice to be between 5 and 8 are :

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

∴ No. of favourable outcomes = 11.

P(getting a sum between 5 and 8) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{11}{36}$.

Hence, the probability of getting sum of numbers on the upper-most faces of two dice between 5 and 8 = $\dfrac{11}{36}$.

(iv) Favourable outcomes for sum of numbers on the upper-most faces of two dice to be more than 10 are :

(5, 6), (6, 5), (6, 6)

∴ No. of favourable outcomes = 3.

P(getting a sum of more than 10) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{36} = \dfrac{1}{12}$.

Hence, the probability of getting sum of numbers on the upper-most faces of more than 10 = $\dfrac{1}{12}$.

(v) Favourable outcomes for sum of numbers on the upper-most faces of two dice to be less than 6 are :

(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1).

∴ No. of favourable outcomes = 10.

P(getting a sum of less than 6) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{10}{36} = \dfrac{5}{18}$.

Hence, the probability of getting sum of numbers on the upper-most faces of less than 6 = $\dfrac{5}{18}$.