In a single throw of two dice, what is the probability of getting

(i) a total of 9

(ii) two aces (ones)

(iii) at least one ace

(iv) a doublet

(v) five on one die and six on the other

(vi) a multiple of 2 on one and a multiple of 3 on the other ?

In a single throw of two dice.

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

(i) Favourable outcomes for getting a total of 9 = {(3, 6), (4, 5), (5, 4), (6, 3)}.

∴ No. of favourable outcomes = 4.

P(of getting a total of 9) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{4}{36} = \dfrac{1}{9}$.

Hence, probability of getting a total of 9 = $\dfrac{1}{9}$.

(ii) Favourable outcomes for getting two aces = {(1, 1)}.

∴ No. of favourable outcomes = 1.

P(of getting two aces) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{1}{36}$.

Hence, probability of getting two aces = $\dfrac{1}{36}$.

(iii) Favourable outcomes for getting at least one ace = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1)}.

∴ No. of favourable outcomes = 11.

P(of getting at least one ace) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{11}{36}$.

Hence, probability of getting at least one ace = $\dfrac{11}{36}$.

(iv) Favourable outcomes for getting a doublet = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}.

∴ No. of favourable outcomes = 6.

P(of getting a doublet) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{6}{36} = \dfrac{1}{6}$.

Hence, probability of getting a doublet = $\dfrac{1}{6}$.

(v) Favourable outcomes for getting 5 on one die and 6 on other = {(5, 6), (6, 5)}.

∴ No. of favourable outcomes = 2.

P(of getting 5 on one die and 6 on other)

= $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{36} = \dfrac{1}{18}$.

Hence, probability of getting 5 on one die and 6 on other = $\dfrac{1}{18}$.

(vi) Favourable outcomes for getting a multiple of 2 on one die and a multiple of 3 on other = {(2, 3), (2, 6), (3, 2), (3, 4), (3, 6), (4, 3), (4, 6), (6, 2), (6, 3), (6, 4), (6, 6)}.

∴ No. of favourable outcomes = 11.

P(of getting a multiple of 2 on one die and a multiple of 3 on other)

= $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{11}{36}$.

Hence, probability of getting a multiple of 2 on one die and a multiple of 3 on other = $\dfrac{11}{36}$.