Question

Given, two independent events $A$ and $B$ such that $P(A)=0.3,P(B)=0.6$. Then, match the terms of column I with their respective values in column II and choose the correct option from the codes given below.

Column I		Column II
A	$P(A$ and $B)$	1	0.28
B	$P(A$ and $\text{not}B)$	2	0.72
C	$P(A$ or $B)$	3	0.12
D	P(neither $A$ nor $B$ )	4	0.18

• A. A- 4 ,B- 3 ,C- 1 ,D- 2
• B. A- 4 ,B-2,C- 1 ,D- 3
• C. A- 1 ,B- 2 ,C- 3 ,D- 4
• D. A- 4 ,B-3,C- 2 ,D- 1

Answer 1

Answer: (D)

If $A$ and $B$ are two independent events, then $A^{\prime }$ and $B,A$ and $B^{\prime },A^{\prime }$ and $B^{\prime }$ are all independent events.
It is given that $P(A)=0.3$ and $P(B)=0.6$
Also, $A$ and $B$ are independent events.
A. $P(A$ and $B)=P(A\cap B)=P(A)\times P(B)$
$=0.3\times 0.6=0.18(\because A$ and $B$ are independent)
B. $P(A$ and not $B)=P\left(A\cap B^{\prime }\right)=P(A)\times P\left(B^{\prime }\right)$
$\left(\begin{array}{c}\because A\text{ and }B\text{ are independent, }\\ \therefore A\text{ and }{B}^{\prime }\text{ are also independent }\end{array}\right)$
$=(0.3)[1-P(B)]$
$=(0.3)(1-0.6)$
$=0.3\times 0.4=0.12$
C. $P(A\text{ or }B)=P(A\cup B)$
$=P(A)+P(B)-P(A\cap B)=P(A)+P(B)-P(A)\times P(B)$
$=0.3+0.6-0.3\times 0.6=0.9-0.18=0.72$
D. $P$ (neither $A$ nor $B)=P(A^{\prime }$ and $B^{\prime })$
$=P\left(A^{\prime }\cap B^{\prime }\right)=P(A\cup B{)}^{\prime }$
$\left[\because P\left(A^{\prime }\cap B^{\prime }\right)=P(A\cup B{)}^{\prime }\right]$
$=1-P(A\cup B)$
$-1-0.72-028$ [using part $(C)]$
Alternate Method
$P(\text{ neither }A\text{ nor }B)=P\left(A^{\prime }\cap B^{\prime }\right)=P\left(A^{\prime }\right)P\left(B^{\prime }\right)$
$\left(\begin{array}{c}\therefore A\text{ and }B\text{ are independent }\\ \therefore A^{\prime }\text{ and }{B}^{\prime }\text{ are also independent }\end{array}\right)$
$=[1-P(A)][1-P(B)]$
$=(1-0.3)(1-0.6)=0.7\times 0.4=0.28$