# Theorem of Total Probability (Addition Theorem)

## Statement

If A and B are two events with their respective probabilities P(A) and P(B), then the probability of occurrence of at least one of these two events denoted by P(A∪B) is given by;

P(A∪B) = P(A) + P(B) – P(A∩B)

where, P(A∩B) is the probability of the simultaneous occurrence of the events A and B.

## Proof

Let *n *be the total number of equally likely and exhaustive outcomes (or cases) of a random experiment. If *u, v *and* w *be the number of cases favorable to the events A, B and A∩B (common to A and B) respectively, then

P(A) = u/n

P(B) = v/n

P(A∩B) = w/n

Since the outcomes *w * are common to both events A and B, so the cases favorable to the event A∪B are (u+v-w).

Now,

P(A∪B) = (u+v-w)/n = u/n + v/n – w/n

i.e. P(A∪B) = P(A) + P(B) – P(A∩B)

### Case I:

If A and B are two mutually exclusive events, then,

P(A∪B) = P(A) + P(B)

### Case II:

If A, B and C are three events, then the probability of occurence of at least one of these events is,

P(A∪B∪C) = P(A) + P(B) + P(C) – P(A∩B) – P(B∩C) – P(A∩C) + P(A∩B∩C)

*This question has been asked for 4 marks in the Grade 12 final examination Mathematics.*