# 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

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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