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).

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

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

P (B/A) = n_{2}/n_{1}

Therefore,

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

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).

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