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Addition Theorem

If A and B are two events associated with a random experiment, then P(A ∪ B) = P(A) + P(B) – P(A  B)
Corollary: If the events are mutually exclusive, then P(A  B) = P(A) + P(B)
Addition theorem can be extended for any number of events.
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?
Out of 30 items, two can be selected in 30C2 ways. So, exhausted number of cases = 30C2.
Consider the following events:
A = getting two apples; B = getting two good items
Required probability = P (A∪B) = P(A) + P(B) – P(A∩B)
There are 20 apples, out of which two can be drawn in 20C2 ways.
∴ P(A) = Description: 1599.png
There are 8 defective pieces and the remaining 22 are good. Out of 22 good pieces, two can be selected in 22C2 ways.
∴ P(B) = Description: 1608.png
Since there are 15 pieces which are good apples of which 2 can be selected in 15C2 ways, therefore
P (A ∩ B) = Probability of getting 2 pieces which are good apples = Description: 1617.png
From (i)
Required probability = P(A) + P(B) – P(A∩B)
Description: 1626.png

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