If A and B are two events such that P(A) = fr | vedclass

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(A) We are given $P(A) = \frac{1}{4}$,$P(B) = \frac{1}{2}$,and $P(A \cap B) = \frac{1}{8}$.We need to find $P(A' \cap B')$,where $A'$ and $B'$ are the

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$\frac{3}{8}$

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(A) We are given $P(A) = \frac{1}{4}$,$P(B) = \frac{1}{2}$,and $P(A \cap B) = \frac{1}{8}$.We need to find $P(A' \cap B')$,where $A'$ and $B'$ are the complements of events $A$ and $B$ respectively.By De Morgan's Law,$A' \cap B' = (A \cup B)'$.Therefore,$P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B)$.Using the addition theorem,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.$P(A \cup B) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8} = \frac{2+4-1}{8} = \frac{5}{8}$.Thus,$P(A' \cap B') = 1 - \frac{5}{8} = \frac{3}{8}$.

Box	White	Black
$B_1$	$1$	$2$
$B_2$	$3$	$1$
$B_3$	$2$	$3$

If $A$ and $B$ are two events such that $P(A) = \frac{1}{4}$,$P(B) = \frac{1}{2}$,and $P(A \cap B) = \frac{1}{8}$,find $P(\text{not } A \text{ and not } B)$.

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