Given that the two numbers appeaing on throwing two dices are different. Find the probability o the event ‘the sum of numbers on the dice is 4’.

.: When a pair of dice is rolled once, then the sample space is

(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)

Let $E$ . "the sum of the number on the dice is 4'

and $F:$ ‘number appearing on the two dice are different'

then $E = \{ (1,3),(2,2),(3,1)\} \Rightarrow P(E) = 3/36$

$F$ contains all points of $S$ except

(1,1),(2,2),(3,3),(4,4),(5,5),(6,6) .

This means that $F$ contains $36 - 6 = 30$ sample points

$\Rightarrow$ $P(F) = \cfrac{{30}}{{36}}$

$\Rightarrow$ $E \cap F = \{ (1,3),(3,1)\} \Rightarrow P(E \cap F) = \cfrac{2}{{36}}$

Hence, the required probability

$= P(E|F) = \cfrac{{P(E \cap F)}}{{P(F)}} = \cfrac{{2/36}}{{30/36}} = \cfrac{2}{{30}} = \cfrac{1}{{15}}$