If P(A) = frac{1}{2},P(B) = frac{1}{3} and P( | vedclass

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(B) By De Morgan's Law,$P(A' \cap B') = P((A \cup B)')$.
$P(A' \cap B') = 1 - P(A \cup B)$.
Using the addition theorem of pr

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$\frac{5}{12}$

Vedclass · Answer

(B) By De Morgan's Law,$P(A' \cap B') = P((A \cup B)')$.
$P(A' \cap B') = 1 - P(A \cup B)$.
Using the addition theorem of probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $P(A \cup B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{4}$.
Finding a common denominator of $12$: $P(A \cup B) = \frac{6}{12} + \frac{4}{12} - \frac{3}{12} = \frac{7}{12}$.
Therefore,$P(A' \cap B') = 1 - \frac{7}{12} = \frac{5}{12}$.

If $P(A) = \frac{1}{2}$,$P(B) = \frac{1}{3}$ and $P(A \cap B) = \frac{1}{4}$,then the value of $P(A' \cap B')$ is

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