Considering the theory of Independent event, if we assume A and B are two separate independent events in a sample space of S and if these two facts are true as:

• P (A) ≠ 0
• P (B) 0

Then the probabi1ity of happening of both the events are given by

• P (AB) = P (A).P (B/A)
• P (AB) = P (B).P (A/B)

Note: Here P(B/A) is the conditional probability.This is simply to provide information on event A which has already occurred. Without A appearing beforehand event B cannot occur.  Same goes for P (A/B).

• Let us supposen is the possible outcomes from a random experiments.The results are almost equivalent. So, if n1 of these outcomes is favorable to the event A, then we’ll come to an equation of:

P (A) = n1/n

• And if n2outcomes be favorable to another event B, then the equation is:
• P (AB) = n2/n
• When we are mentioning n2 then it is the number of outcomes favorable both in event A and B.

Applying the theory of conditional probability of event A we come to:

P (B/A) = n2/n1

Therefore,

n2/n = n1/n* n2/n1

P (AB) = P (A). P (B/A)

If another event, suppose C is occurring with other two previous events A and B then we’ll get:

P (ABC) = P (A). P (B/A) P(C/AB).  