Let x denote the outcome on black die and y denote the outcome on red die, then sample space is
S = {(x, y) : x, y ∈ (1, 2, 3, 4, 5, 6)}, which contain 6 × 6 = 36 equally likely simple events.
(a) E : ‘sum greater than 9’ and F : ‘black die resultedin a 5’,
i.e., E = {(6, 4), (4, 6), (5, 5), (5, 6), (6, 5), (6, 6)}
and F = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
⇒ E ∩ F = {(5, 5), (5, 6)}
P(E) =

6

36

, P(F) =

6

36

, P(E∩F) =

2

36

Required probability =P(E|F)=

P(E∩F)

P(F)

=

2 36

6 36

=

2

6

=

1

3

(b) E : ‘a total of 8’ and F : ‘red die resulted in a number less than 4’
i.e., E = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
and F = {(x, y) : x ∈ {1, 2, 3, 4, 5, 6}, y ∈ {1, 2, 3}}
i.e., F = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (6, 1), (6, 2), (6, 3)}
Hence, E ∩ F = {(5, 3), (6, 2)}, P(E) =

5

36

, P(F) =

18

36

, P(E∩F) =

2

36

∴ Required probability =P(E|F)=

P(E∩F)

P(F)

=

2 36

18 36

=

2

18

=

1

9