If E and F are events such that P(E)=frac{1}{ | vedclass

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(A) Given: $P(E) = \frac{1}{4}$,$P(F) = \frac{1}{2}$,and $P(E \cap F) = \frac{1}{8}$.Using the addition theorem of probability,$P(E \cup F) = P(E) + P

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$\frac{3}{8}$

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(A) Given: $P(E) = \frac{1}{4}$,$P(F) = \frac{1}{2}$,and $P(E \cap F) = \frac{1}{8}$.Using the addition theorem of probability,$P(E \cup F) = P(E) + P(F) - P(E \cap F)$.$P(E \cup F) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8} = \frac{2+4-1}{8} = \frac{5}{8}$.We need to find $P(	ext{not } E 	ext{ and not } F)$,which is $P(E' \cap F')$.By De Morgan's Law,$E' \cap F' = (E \cup F)'$.Therefore,$P(E' \cap F') = P((E \cup F)') = 1 - P(E \cup F)$.$P(E' \cap F') = 1 - \frac{5}{8} = \frac{3}{8}$.

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

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