Let the first observation be from the black die and second from the red die. 

When two dice (one black and another red) are rolled, the sample space 

S = 6 x 6 = 36 (equally likely sample events) 

(i) Let E : set of events in which sum greater than 9 and F : set of events in which black die resulted in a 5 

E = {(6,4), (4,6), (5, 5), (5,6), (6, 5), (6,6)} ⇒ n(E) = 6 

and F = {(5, 1), (5,2), (5, 3), (5, 4), (5, 5), (5,6)} ⇒ n(F) = 6 

⇒ E ∩ F = {(5, 5), (5,6)} ⇒ n(E ∩ F) = 2

The conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5, is given by P(E/F)

P(E) = 6/36 and P(F) = 6/36

Also, P(E ∩ F) = 2/36

∴ P(E/F) = (P(E ∩ F))/P(F) = (2/36)/(6/36) = 2/6 = 1/3

(ii) Let E : set of events having 8 as the sum of the observations, 

F : set of events in which red die resulted in a (in any one die) number less than 4 

⇒ E = {(2,6), (3, 5), (4,4), (5, 3), (6,2)} ⇒n(E) = 5 

and F = {(1, 1), (1, 2),….. (3, 1), (3, 2)…… (5, 1), (5, 2),……….} ⇒ n(F) = 18 

⇒ E ∩ F = {(5, 3), (6, 2)} ⇒ n(E ∩ F) = 2 

The conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5, is given by P(E/F)

P(E) = 5/36 and P(F) = 18/36

Also, P(E ∩ F) = 2/36

∴ P(E/F) = (P(E ∩ F))/P(F) = (2/36)/(18/36) = 2/18 = 1/9