In a single throw of two dice, find the probability of :

(i) a doublet

(ii) a number less than 3 on each dice

(iii) an odd number as a sum

(iv) a total of atmost 10

(v) an odd number on one dice and a number less than or equal to 4 on other dice.

When two dice are rolled simultaneously;

Total number of possible outcomes = 6 × 6 = 36.

(i) For obtaining a doublet, the favourable outcomes are : (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6).

∴ Number of favourable outcomes = 6

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

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

(ii) For obtaining a number less than 3 on each dice, the favourable outcomes are : (1, 1), (1, 2), (2, 1), (2, 2).

∴ Number of favourable outcomes = 4

P(getting a number less than 3 on each dice) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{4}{36} = \dfrac{1}{9}$.

Hence, the probability of getting a number less than 3 on each dice = $\dfrac{1}{9}$.

(iii) For obtaining an odd number as a sum, the favourable outcomes are : (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5).

∴ Number of favourable outcomes = 18

P(getting an odd number as a sum) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{18}{36} = \dfrac{1}{2}$.

Hence, the probability of getting an odd number as the sum = $\dfrac{1}{2}$.

(iv) For obtaining a total of atmost 10, the favourable outcomes 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), (6, 1), (6, 2), (6, 3), (6, 4).

∴ Number of favourable outcomes = 33

P(getting a total of atmost 10) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{33}{36} = \dfrac{11}{12}$.

Hence, the probability of getting a total of atmost 10 = $\dfrac{11}{12}$.

(v) For obtaining an odd number on one dice and a number less than or equal to 4 on other dice, the favourable outcomes are : (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 3), (2, 5), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (4, 1), (4, 3), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4).

∴ Number of favourable outcomes = 20

P(getting an odd number on one dice and a number less than or equal to 4 on other dice) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{20}{36} = \dfrac{5}{9}$.

Hence, the probability of getting an odd number and a number less than or equal to 4 on other dice = $\dfrac{5}{9}$.