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Theorem of Compound Probability/Multiplication Theorem

Let A and B be two events in a sample space S and P (A) ≠ 0 P (B) 0, then the probabi1ity of happening of both the events are given by

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

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

[Here P(BIA) denote the conditional probability which means the probability of event B such that the event A has already occurred otherwise event B will not occur. Similarly P (A/B).]

            Proof: Let n possible outcomes from a random experiment which are mutually exclusive, exhaustive and equally likely. If n 1 of these outcomes is favorable to the event A, the unconditional probability of A is

 

P (A) = n1/n

 

Out of these n 1 outcomes, let n2 outcomes be favorable to another event B i.e., the number

Of outcomes favorable to A as well as B is n2. Hence,

P (AB) = n2/n

 

Then the conditional probability of B assuming that A has already occurred is

P (B/A) = n2/n1

 

Therefore,

n2/n = n1/n* n2/n1

P (AB) = P (A). P (BIA) which is (i).

 

Similarly, we can prove (ii).

Note. If the occurrence of the event A as well as B as well as C is given by

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