$\begin{array}{1 1} \text{A and B are independent} \\\text{A and B are not independent} \end{array} $

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- If A and B are independent events, \(P(A\cap\;B)=P(A)\;P(B)\)
- P (not A and not B) = \(P(\bar{A}\cap\;\bar{B})\) = \(P(\overline{A \cup B})\) = 1 - P (A $\cup$ B)
- P (A $\cup$ B) = P(A) + P(B) - P(A $\cap$ B)
- P (\(\bar{A})\) = 1 - P(A)

Given \(\;P(A)=\large \frac{1}{2}, \) \(\;;P(B)=\large \frac{7}{12}, \)\(\;\) and \(\;P(\bar{A}\cup\;\bar{B})=\large \frac{1}{4}\)

If A and B are independent events, \(P(A\cap\;B)=P(A)\;P(B)\)

We know that P (not A and not B) = \(P(\bar{A}\cap\;\bar{B})\) = \(P(\overline{A \cup B})\) = 1 - P (A $\cup$ B)

$\Rightarrow \large \frac{1}{4} $$= 1 - P (A \cup B) \rightarrow P (A \cup B) = 1 - \large \frac{1}{4} = \frac{3}{4}$

$\Rightarrow P (A \cap B) = P (A) + P(B) - P (A \cup B) = \large \frac{1}{2} + \frac{7}{12} - \frac{3}{4}$ = $\large \frac{6 + 7 - 4}{12} = \frac{9}{12} = \frac{3}{4}$

However, P(A) $\times$ P(B) = $\large \frac{1}{2}$ $\times$ $\large\frac{7}{12} = \frac{7}{24} $$\;\neq P (A \cap B)$

Therefore A and B are not independent.

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