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 $B)=P\left(A \cap B^{\prime}\right)=P(A) \times P\left(B^{\prime}\right)$
$\begin{pmatrix}\because A \text { and } B \text { are independent, } \\\therefore A \text { and } B^{\prime} \text { are also independent }\end{pmatrix}$
$=(0.3)[1-P(B)]$
$=(0.3)(1-0.6)$
$=0.3 \times 0.4=0.12$
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 $B)=P\left(A^{\prime}\right.$ and $\left.B^{\prime}\right)$
$=P\left(A^{\prime} \cap B^{\prime}\right)=P(A \cup B)^{\prime}$
$\left[\because P\left(A^{\prime} \cap B^{\prime}\right)=P(A \cup B)^{\prime}\right]$
$=1-P(A \cup B)$
$-1-0.72-028 $ [using part $(C)]$
Alternate Method
$P(\text { neither } A \text { nor } B)=P\left(A^{\prime} \cap B^{\prime}\right)=P\left(A^{\prime}\right) P\left(B^{\prime}\right)$
$\begin{pmatrix}\therefore A \text { and } B \text { are independent } \\ \therefore A^{\prime} \text { and } B^{\prime} \text { are also independent }\end{pmatrix}$
$=[1-P(A)][1-P(B)]$
$=(1-0.3)(1-0.6)=0.7 \times 0.4=0.28$