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Let A and B be the events such that P(A) = 1/2, P(B) = 7/12 and P(not A or not B) = 1/4.

State whether A and B are

(i) mutually exclusive

(ii) independent

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Given: A and B are the events such that P(A) = \(\frac{1}{2}\) and P(B) = \(\frac{7}{12}\) and 

P(not A or not B) = \(\frac{1}{4}\)

To Find: 

(i) If A and B are mutually exclusive

Since P(not A or not B) = \(\frac{1}{4}\) i.e., P\((\overline A\cup\overline B)\) = \(\frac{1}{4}\)

we know that , P\((\overline A\cup\overline B)\)) = P(A ∩ B)' = 1 - P(A ∩ B) = 0

But here P(A ∩ B) \(\neq\) 0

Therefore , A and B are not mutually exclusive.

(ii) If A and B are independent

The condition for two events to be independent is given by

P(E1 ∩ E2) = P(E1) x P(E2)

\(\frac{1}{2}\times \frac{7}{12}\)

\(\frac{7}{24}\) Equation 2

Since Equation 1 \(\neq\) Equation 2

\(\Rightarrow\) A and B are not independent

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