A black and a red dice are rolled.

(a) Find the conditional probability of obtaining a sum greater than 9, given that the black dice resulted in a 5.

(b) Find the conditional probability of obtaining the sum 8, given that the red dice resulted in a number less than 4.

.: Let $x$ denote the outcome on black dice and $y$

denote the outcome on red dice,

then sample space is
$S = \{ (x,y):x,y \in (1,2,3,4,5,6)\} ,$ which contain $6 \times 6 = 36$ equally likely simple-events.

(a) Let $E:$ 'sum greater than ${9^\prime }$ and $F:$ 'black dice resulted in a 5 ',

i.e. $E = \{ (6,4),(4,6),(5,5),(5,6),(6,5),(6,6))$

and $F = \{ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\}$

$\Rightarrow E \cap F = \{ (5,5),(5,6)\}$

$P(E) = 6/36,P(F) = 6/36,P(E \cap F) = 2/36$

Required probability $= P(E/F) = \cfrac{{P(E \cap F)}}{{P(F)}} = \cfrac{{2/36}}{{6/36}} = \cfrac{2}{6} = \cfrac{1}{3}$

(b) Let $E:$ 'a total of 8 ' and $F$ : 'red dice resulted in a number less than 4 '

i.e., $E = \{ (2,6),(3,5),(4,4),(5,3),(6.2)\}$
and $F = \{ (x,y):x \in \{ 1,2,3,4,5,6\} ,y \in \{ 1,2,3\} \}$

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, $E \cap F = \{ (5,3),(6,2)\} P(E) = 5/36,P(F) = 18/36,P(E \cap F) = 2/36$

$\therefore$ Required probability
$= P(E|F) = \cfrac{{P(E \cap F)}}{{P(F)}} = \cfrac{{2/36}}{{18/36}} = \cfrac{2}{{18}} = \cfrac{1}{9}$