Two different dice are thrown at the same time. Find the probability of getting :

(i) a doublet

(ii) a sum of 8

(iii) sum divisible by 5

(iv) sum of atleast 11.

(i) Let A be the event of getting 'a doublet', then

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

∴ The number of outcomes favourable to event A = 6.

∴ P(a doublet) = $\dfrac{6}{36} = \dfrac{1}{6}.$

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

(ii) Let B be the event of getting 'a sum of 8', then

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

∴ The number of outcomes favourable to event B = 5.

∴ P(a sum of 8) = $\dfrac{5}{36}.$

Hence, the probability of getting a sum of 8 is $\dfrac{5}{36}$.

(iii) Let C be the event of getting 'a sum divisible by 5', then

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

∴ The number of outcomes favourable to event C = 7.

∴ P(a sum divisible by 5) = $\dfrac{7}{36}.$

Hence, the probability of getting a sum divisible by 5 is $\dfrac{7}{36}$.

(iv) Let D be the event of getting 'sum of atleast 11', then

D = {(5, 6), (6, 5), (6, 6)}.

∴ The number of outcomes favourable to event D = 3.

∴ P(sum of atleast 11) = $\dfrac{3}{36} = \dfrac{1}{12}.$

Hence, the probability of getting a sum of atleast 11 is $\dfrac{1}{12}$.