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Let $A$ and $B$ be events in a sample space $S$ such that $P(A)=0.5, \quad P(B)=0.4$ and $P(A \cup B)=0.6$. Observe the following lists.
The correct match of List I from List II is
(i) (ii) (iii) (iv)
Options:
The correct match of List I from List II is
(i) (ii) (iii) (iv)
Solution:
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Verified Answer
The correct answer is:
(3) (2) (4) (1)
Given, $P(A)=0.5, P(B)=0.4$
and $\quad P(A \cup B)=0.6$
$\begin{array}{lc}
\text { (i) } & \because P(A \cup B)=P(A)+P(B)-P(A \cap B) \\
\Rightarrow & 0.6=0.5+0.4-P(A \cap B) \\
\Rightarrow & P(A \cap B)=0.3
\end{array}$
(ii)$\begin{aligned}
P(A \cap \bar{B}) & =P(A)-P(A \cap B) \\
& =0.5-0.3=0.2
\end{aligned}$
(iii)$\begin{aligned}
P(\bar{A} \cap B) & =P(B)-P(A \cap B) \\
& =0.4-0.3=0.1
\end{aligned}$
(iv)$\begin{aligned}
& P \overline{(A} \cap \bar{B})=P(A \cup B)^c \\
= & 1-P(A \cup B)=1-0.6 \\
= & 0.4
\end{aligned}$
and $\quad P(A \cup B)=0.6$
$\begin{array}{lc}
\text { (i) } & \because P(A \cup B)=P(A)+P(B)-P(A \cap B) \\
\Rightarrow & 0.6=0.5+0.4-P(A \cap B) \\
\Rightarrow & P(A \cap B)=0.3
\end{array}$
(ii)$\begin{aligned}
P(A \cap \bar{B}) & =P(A)-P(A \cap B) \\
& =0.5-0.3=0.2
\end{aligned}$
(iii)$\begin{aligned}
P(\bar{A} \cap B) & =P(B)-P(A \cap B) \\
& =0.4-0.3=0.1
\end{aligned}$
(iv)$\begin{aligned}
& P \overline{(A} \cap \bar{B})=P(A \cup B)^c \\
= & 1-P(A \cup B)=1-0.6 \\
= & 0.4
\end{aligned}$
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