# Let A and B be two events such that P(A $$\cup$$ B)’ = 1/6, P(A $$\cap$$ B) = 1/4 and P(A)’ = 1/4 where A’ stands for complement of A. Then prove that events A and B independent

## Solution :

Given P(A $$\cup$$ B)’ = 1/6, P(A $$\cap$$ B) = 1/4 and P(A)’ = 1/4

$$\therefore$$   P(A $$\cup$$ B) = 1 – P(A $$\cup$$ B)’

= 1 – $$1\over 6$$ = $$5\over 6$$

and P(A) = 1 – P(A)’ = 1 – $$1\over 4$$ = $$3\over 4$$

P(A $$\cup$$ B) = P(A) + P(B) – P(A $$\cap$$ B)

$$\implies$$ $$5\over 6$$ = $$3\over 4$$ + P(B) – $$1\over 4$$

P(B) = $$1\over 3$$

$$\implies$$ A and B are not equally likely,

Also, P(A $$\cap$$ B) = P(A).P(B) = $$1\over 4$$

So, events are independent.

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