If P(A) = P(B) = x and P(A cap B) = P(A' cap | vedclass

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(A) Given $P(A) = P(B) = x$ and $P(A \cap B) = \frac{1}{3}$.Also,$P(A' \cap B') = \frac{1}{3}$.By De Morgan's Law,$P(A' \cap B') = P((A \cup B)') = 1

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$\frac{1}{2}$

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(A) Given $P(A) = P(B) = x$ and $P(A \cap B) = \frac{1}{3}$.Also,$P(A' \cap B') = \frac{1}{3}$.By De Morgan's Law,$P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B) = \frac{1}{3}$.Therefore,$P(A \cup B) = 1 - \frac{1}{3} = \frac{2}{3}$.Using the addition theorem,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Substituting the values,$\frac{2}{3} = x + x - \frac{1}{3}$.$\frac{2}{3} + \frac{1}{3} = 2x$.$1 = 2x \Rightarrow x = \frac{1}{2}$.

If $P(A) = P(B) = x$ and $P(A \cap B) = P(A' \cap B') = \frac{1}{3}$,then $x = $

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