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$A$ and $B$ are two events such that $P(A)=0.54$, $P(B)=0.69$ and $P(A \cap B)=0.35$.
Find (i) $P(A \cup B)$ (ii) $P\left(A^{\prime} \cap B^{\prime}\right)$
(iii) $P\left(A \cap B^{\prime}\right)$ (iv) $P\left(B \cap A^{\prime}\right)$
Find (i) $P(A \cup B)$ (ii) $P\left(A^{\prime} \cap B^{\prime}\right)$
(iii) $P\left(A \cap B^{\prime}\right)$ (iv) $P\left(B \cap A^{\prime}\right)$
Solution:
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Verified Answer
(i) $\quad P(A \cup B)=P(A)+P(B)-P(A \cap B)$ $=0.54+0.69-0.35=1.23-0.35=0.88$
(ii) $A^{\prime} \cap B^{\prime}=(A \cup B)^{\prime}$ By De-Morgan's Law
$\begin{aligned}
\therefore & P\left(A^{\prime} \cap B^{\prime}\right)=P(A \cup B)^{\prime}=1-P(A \cup B) \\
=1-0.88=0.12
\end{aligned}$
(iii) $A \cap B^{\prime}=A-A \cap B$
$\begin{aligned}
&\therefore P\left(A \cap B^{\prime}\right)=P(A)-P(A \cap B) \\
&=0.54-0.35=0.19
\end{aligned}$
(iv) $P\left(B \cap A^{\prime}\right)=P(B)-P(A \cap B)$
$=0.69-0.35=0.34$
(ii) $A^{\prime} \cap B^{\prime}=(A \cup B)^{\prime}$ By De-Morgan's Law
$\begin{aligned}
\therefore & P\left(A^{\prime} \cap B^{\prime}\right)=P(A \cup B)^{\prime}=1-P(A \cup B) \\
=1-0.88=0.12
\end{aligned}$
(iii) $A \cap B^{\prime}=A-A \cap B$
$\begin{aligned}
&\therefore P\left(A \cap B^{\prime}\right)=P(A)-P(A \cap B) \\
&=0.54-0.35=0.19
\end{aligned}$
(iv) $P\left(B \cap A^{\prime}\right)=P(B)-P(A \cap B)$
$=0.69-0.35=0.34$
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