# Addition Theorem of Probability (for Mutually Exclusive Events)

Let A and B be two mutually exclusive events with respective probabilities P(A) and P(B).

Then, probability of occurrence of at least one of these two events is given by the sum of the individual probabilities.

i.e., P(A  B) = P(A) + P(B)

Example
A number is selected from the first 50 natural numbers. What is the probability that it is a multiple of 5 or 11?
Solution
Let S be the sample space, i.e.S = {1, 2, 3, 4, 5, … 49, 50}
Let A be the set of all the multiples of 5 within the list of first 50 natural numbers,
i.e.A = {5, 10, 15, 20, 25, 30, 35, 40, 45, 50} Let B be the set of all the multiples of 11 within the list of first 50 natural numbers,
i.e.B = {11, 22, 33, 44} There is no common outcome in events A and B, hence the two events are mutually exclusive. The probability that the selected number will be either a multiple of 5 or 11 will be
P(A ∪ B) = P(A) + P(B) = 0.2 + 0.08 = 0.28
We can extend the above concept when there are more than two events.
If A1A2A3 … An are n mutually exclusive events and P(A1), P(A2), P(A3) … P(An) are their respective probabilities, then the probability of the occurrence of any one of the events will be the sum of their individual probabilities.
i.e.P(A1 ∪ A2 ∪ A3 … ∪ An) = P(A1) + P(A2) + P(A3) … + P(An)
The above two cases are applicable only when the events are mutually exclusive.
If events A1 and A2 are not mutually exclusive, i.e., if there is a chance for the two events A1 and A2 to occur simultaneously, then the probability that either the event A1 or A2 will occur is given by the sum of the probability of the individual events minus the probability of both the events occurring simultaneously.
i.e.P(A1 ∪ A2) = P(A1) + P(A2) − P(A1 ∩ A2)  