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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.
 
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 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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