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)

Question

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)

Aicharya · Accepted Answer

Given, $P(A)=0.5, P(B)=0.4$ and $\quad P(A \cup B)=0.6$ $\begin{array}{lc} 	ext { (i) } & \because P(A \cup B)=P(A)+P(B)-P(A \cap B) \ \Rightarrow & 0.6=0.5+0.4-P(A \cap B) \ \Rightarrow & P(A \cap B)=0.3 \end{array}$ (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)$\begin{aligned} & P \overline{(A} \cap \bar{B})=P(A \cup B)^c \ = & 1-P(A \cup B)=1-0.6 \ = & 0.4 \end{aligned}$

Let $A$ and $B$ be events in a sample space $S$ such that $P(A)=0.5, \quad P(B)=0.4$ and $P(A \cup B)=0.6$.…

Solution