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$A$ and $B$ are events such that $P(A)=0.42, P(B)=0.48$ and $P(A$ and $B)=0.16$. Determine
(i) $P($ not $A)$,
(ii) $P($ not $B)$ and (iii) $P(A$ or $B)$
(i) $P($ not $A)$,
(ii) $P($ not $B)$ and (iii) $P(A$ or $B)$
Solution:
1555 Upvotes
Verified Answer
(i) $\quad P(\operatorname{not} A)$
$=P\left(A^{\prime}\right)=1-P(A)=1-0.42=0.58$
(ii) $P($ not $B)=P\left(B^{\prime}\right)=1-P(B)$
$=1-0.48=0.52$
(iii) $P(A$ or $B)$
$\begin{aligned}
&=P(A \cup B)=P(A)+P(B)-P(A \cap B) \\
&=0.42+0.48-0.16=0.74
\end{aligned}$
$=P\left(A^{\prime}\right)=1-P(A)=1-0.42=0.58$
(ii) $P($ not $B)=P\left(B^{\prime}\right)=1-P(B)$
$=1-0.48=0.52$
(iii) $P(A$ or $B)$
$\begin{aligned}
&=P(A \cup B)=P(A)+P(B)-P(A \cap B) \\
&=0.42+0.48-0.16=0.74
\end{aligned}$
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