We will solve this Problem using Set Theoretic Approach of
Probability.
An ordered pair #(i,j)# denotes the outcome of getting a no. #i# in
the first roll and #j# in the second one of the cube, #1lei,jle6.#
The Sample Space, #U#, associated with the random experiment of
rolling a no. cube, is,
#U={(1,1),(1,2),...,(1,6),(2,1),(2,2),...,(2,6),...,(6,1),(6,2),...(6,6)}.#
#:. n(U)"=the no. of elements in "U=36.#
Let #A=# The Event that the sum of two rolls is less than #6#.
#:. A={(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)}.#
Let #B=# The Event that the first roll is no.#3#.
#:. B={(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)}.#
#rArr n(B)=6 rArr P(B)=(n(B))/(n(U))=6/36.#
Now, the Reqd. Prob #=P(A/B)=(P(AnnB))/(P(B))...[Defn.]#
But, #AnnB={(3,1),(3,2)} rArr P(AnnB)=2/36#.
#:. P(A/B)=(2/36)/(6/36)=2/6=1/3#.
Enjoy Maths.!
