Question

Let $A$ and $B$ be events in a sample space $S$ such that $P\left(A\right)=0.5,P\left(B\right)=0.4$ and $P(A\cup B)=0.6$. Observe the following lists:

	List I		List II
i	$P(A\cap B)$	a	0.4
ii	$P(A\cap \overline{B})$	b	0.2
iii	$P(\overline{A}\cap B)$	c	0.3
iv	$P(\overline{A}\cap \overline{B})$	d	0.1

The correct match of List I from List II is

• A. i- a, ii- b, iii- c, iv- d
• B. i- c, ii- b, iii- d, iv- a
• C. i- c, ii- b, iii- a, iv- d
• D. i- c, ii- a, iii- b, iv- d

Answer 1

The correct option is B i- c, ii- b, iii- d, iv- a

Given : $P\left(A\right)=0.5,P\left(B\right)=0.4,P(A\cup B)=0.6$

i. $P(A\cap B)=P\left(A\right)+P\left(B\right)-P(A\cup B)=0.5+0.4-0.6=0.3$

ii. $P(A\cap \overline{B})=P\left(A\right)-P(A\cap B)=0.5-0.3=0.2$

iii. $P(\overline{A}\cap B)=P\left(B\right)-P(A\cap B)=0.4-0.3=0.1$

iv $P(\overline{A}\cap \overline{B})=1-P(A\cup B)=1-0.6=0.4$