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If $E$ and $F$ are events such that $P(E)=\frac{1}{4}, P(F)=\frac{1}{2}$ and $P(E$ and $F)=\frac{1}{8}$, find
(i) $P(E$ or $F)$,
(ii) $P(\operatorname{not} E$ and not $F$ ).
(i) $P(E$ or $F)$,
(ii) $P(\operatorname{not} E$ and not $F$ ).
Solution:
2814 Upvotes
Verified Answer
(i)
$\begin{aligned}
&P(E \text { or } F)=P(E \cup F) \\
&=P(E)+P(F)-P(E \cap F) \\
&=\frac{1}{4}+\frac{1}{2}-\frac{1}{8}=\frac{2+4-1}{8}=\frac{5}{8}
\end{aligned}$
(ii) not $E$ and not $F=E^{\prime} \cap F^{\prime}=(E \cup F)^{\prime}$
$\therefore P($ not $E$ and not $F)$
$=P(E \cup F)^{\prime}=1-P(E \cup F)=1-\frac{5}{8}=\frac{3}{8}$
$\begin{aligned}
&P(E \text { or } F)=P(E \cup F) \\
&=P(E)+P(F)-P(E \cap F) \\
&=\frac{1}{4}+\frac{1}{2}-\frac{1}{8}=\frac{2+4-1}{8}=\frac{5}{8}
\end{aligned}$
(ii) not $E$ and not $F=E^{\prime} \cap F^{\prime}=(E \cup F)^{\prime}$
$\therefore P($ not $E$ and not $F)$
$=P(E \cup F)^{\prime}=1-P(E \cup F)=1-\frac{5}{8}=\frac{3}{8}$
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