1. If two events A and B are mutually exclusive, then the occurrence of either A or B is given by

P (A + B) = P (A) + P (B)

[Also we can write P (A + B) = P (A v B)].

** Proof: **Let *n *possible outcomes from a random experiment which are mutually exclusive, exhaustive and equally likely. If n 1 of these outcomes are favorable to the event A, and *n2 *outcomes are favorable to the event B, then

Since the events A and B are mutually exclusive *i.e., *the n 1 outcomes are completely distinct from *n2 *outcomes, then the number of outcomes favorable to either A or B is n1 + *n2.*

II. When the two events A and B are not mutually exclusive, then the probability of occurrence of at least one of the 2 events is given by

P (A +B) = P (A) + P (B) - P (AB)

Proof: Here the event A + B means the occurrence of one of the following mutually exclusive events: AB, AB and AB. Therefore

P (A +B) = P (AB + AB + AB) = P (AB) + P (AB) + P (AB)

Again, we have P (A) = P (AB) + P (AB)

=> P (AB) = P (A) – P (AB)

And P (B) = P (AB) + P (AB)

=> P (AB) = P (B) – P (AB)

Hence,

P(A + B)= P(AB) +[P(A)- P(AB)]+[P(B)- P(AB)]

= P (A) + P (B) - P (AB).

Note. l. For the events A, B and C wh1ch may not be mutually exclusive,

P (A + B+ C) = P (A) + P (B) + P(C) - P (AB)-P (AC) - P (BC) + P (ABC)

2. Boole's inequality.

P (A +B) ≤ P (A) + P (B).

Here equality sign holds when P (AB) = 0, i ·e., A and B are mutually exclusive.

3. Bonferroni's inequality.

P (AB) ≥ P (A) + P (B) - I.

* *