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 suppose
*n*is the possible outcomes from a random experiments.The results are almost equivalent. So, if n_{1}of these outcomes is favorable to the event A, then we’ll come to an equation of:

P (A) = n_{1}/n

- And if n
_{2}outcomes be favorable to another event B, then the equation is: - P (AB) = n
_{2}/n - When we are mentioning n
_{2}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) = n_{2}/n_{1}

Therefore,

n_{2}/n = n_{1}/n* n_{2}/n_{1}

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).

**Links of Previous Main Topic:-**

- Introduction to statistics
- Knowledge of central tendency or location
- Definition of dispersion
- Moments
- Bivariate distribution
- Theorem of total probability addition theorem

**Links of Next Statistics Topics:-**

- Random variable
- Binomial distribution
- What is sampling
- Estimation
- Statistical hypothesis and related terms
- Analysis of variance introduction
- Definition of stochastic process
- Introduction operations research
- Introduction and mathematical formulation in transportation problems
- Introduction and mathematical formulation
- Queuing theory introduction
- Inventory control introduction
- Simulation introduction
- Time calculations in network
- Introduction of game theory