Two dice are thrown simultaneously. Find the probability of getting:

(i) 10 as the sum of two numbers that turn up

(ii) a doublet of even numbers

(iii) a total of at least 10

(iv) a multiple of 3 as the sum of two numbers that turn up

(i) Let A be the event of 10 as the sum of two numbers that turn up, then

A = {(4, 6), (5, 5), (6, 4)}.

∴ The number of favourable outcomes to the event A = 3.

∴ P(A) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{3}{36} = \dfrac{1}{12}.$

Hence, the probability of 10 as the sum of two numbers that turn up is $\dfrac{1}{12}.$

(ii) Let B be the event of a doublet of even numbers, then

B = {(2, 2), (4, 4), (6, 6)}.

∴ The number of favourable outcomes to the event B = 3.

∴ P(B) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{3}{36} = \dfrac{1}{12}.$

Hence, the probability of getting a doublet of even numbers is $\dfrac{1}{12}.$

(iii) Let C be the event of a total of at least 10, then

C = {(4, 6), (5, 5) , (6, 4), (5, 6), (6, 5), (6, 6)}.

∴ The number of favourable outcomes to the event C = 6.

∴ P(C) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{6}{36} = \dfrac{1}{6}.$

Hence, the probability of getting a total of at least 10 is $\dfrac{1}{6}.$

(iv) Let D be the event of a multiple of 3 as the sum of two numbers that turn up , then

D = {(1, 2), (2, 1), (1, 5), (2, 4), (3, 3) , (4, 2), (5, 1), (3, 6), (4, 5), (5, 4), (6, 3), (6, 6)}.

∴ The number of favourable outcomes to the event D = 12.

∴ P(D) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{12}{36} = \dfrac{1}{3}.$

Hence, the probability of getting a multiple of 3 as the sum of two numbers that turn up is $\dfrac{1}{3}.$