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).
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
