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

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Given, $P(E)=\frac{1}{4}$ , $P(F)=\frac{1}{2}$ and $P(E\hspace{1mm}and\hspace{1mm}F)=\frac{1}{8}$

To find : $P(not\ E\ and\ not\ F) = P(E' \cap F')$

We know,

$P(A' \cap B') = P(A \cup B)' = 1 - P(A \cup B)$

And $P(A\cup B) = P(A)+ P(B) - P(A \cap B)$

$\implies$ $P(E \cup F) =$ $\frac{1}{4}+\frac{1}{2} -\frac{1}{8} = \frac{2+4-1}{8}$

$= \frac{5}{8}$

$\implies$ $P(E' \cap F') = 1 - P(E \cup F)$

$= 1- \frac{5}{8}= \frac{3}{8}$

Therefore, $P(E' \cap F') =$ $\frac{3}{8}$

Posted by

HARSH KANKARIA

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