# Addition Theorem

If A and B are two events associated with a random experiment, then P(A âˆª B) = P(A) + P(B) â€“ P(A âˆ© B)**If the events are mutually exclusive, then P(A âˆª B) = P(A) + P(B)**

*Corollary:*Addition theorem can be extended for any number of events.

Example

A basket contains 20 apples and 10 oranges out of which 5 apples and 3 oranges are defective. If a person takes out 2 at random what is the probability that either both are apples or both are good?

Solution

Out of 30 items, two can be selected in

^{30}C_{2}ways. So, exhausted number of cases =^{30}C_{2}.Consider the following events:

A = getting two apples; B = getting two good items

Required probability = P (AâˆªB) = P(A) + P(B) â€“ P(Aâˆ©B)

There are 20 apples, out of which two can be drawn in

^{20}C_{2}ways.âˆ´ P(A) =

There are 8 defective pieces and the remaining 22 are good. Out of 22 good pieces, two can be selected in

^{22}C_{2}ways.âˆ´ P(B) =

Since there are 15 pieces which are good apples of which 2 can be selected in

^{15}C_{2}ways, thereforeP (A âˆ© B) = Probability of getting 2 pieces which are good apples =

From (i)

Required probability = P(A) + P(B) â€“ P(Aâˆ©B)