$A\text{and}B$ are two events such that $P\left(A\right)=0.54,P\left(B\right)=0.69\text{and}P(A\cap B)=0.35$.
Find
$\left(i\right)$ $P(A\cup B)$
$\left(ii\right)$ $P(A^{\prime }\cap B^{\prime })$
$\left(iii\right)$ $P(A\cap B^{\prime })$
$\left(iv\right)$ $P(B\cap A^{\prime })$

$\left(i\right)$ Calculate $P(A\cup B)$
Given:
$P\left(A\right)=0.54,P\left(B\right)=0.69,P(A\cap B)=0.35$ $...\left(i\right)$
As we know that,
$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$ $....\left(ii\right)$
Putting values of $\left(i\right)$ in $\left(ii\right)$
$P(A\cup B)=0.54+0.69-0.35=0.88$
Hence $P(A\cup B)=0.88$


$\left(ii\right)$ Calculate $P(A^{\prime }\cap B^{\prime })$
$P(A^{\prime }\cap B^{\prime })=P(A\cup B{)}^{\prime }$ $($De Morgan's law$)$
As we know that,
$P(A\cup B{)}^{\prime }+P(A\cup B)=1$
$\Rightarrow P(A\cup B{)}^{\prime }=1-P(A\cup B)$
$\Rightarrow P(A\cup B{)}^{\prime }=1-0.88=0.12$
Hence $P(A\cup B{)}^{\prime }=P(A^{\prime }\cap B^{\prime })=0.12$


$\left(iii\right)$ Calculate $P(A\cap B^{\prime })$
Given :
$P\left(A\right)=0.54,P\left(B\right)=0.69,P(A\cap B)=0.35$ $...\left(i\right)$
From figure below,
$P(A\cap B^{\prime })=P\left(A\right)-P(A\cap B)$ $...\left(ii\right)$
Putting values $\left(i\right)$ in $\left(ii\right)$ then we get
$P(A\cap B^{\prime })=0.54-0.35=0.19$
Hence $P(A\cap B^{\prime })=0.19$
$\left(iv\right)$ Calculate $P(B\cap A^{\prime })$
Given :
$P\left(A\right)=0.54,P\left(B\right)=0.69,P(A\cap B)=0.35$ $...\left(i\right)$
From the figure,
$P(B\cap A^{\prime })=P\left(B\right)-P(A\cap B)$ $...\left(ii\right)$
Putting values $\left(i\right)$ in $\left(ii\right)$ then we get
$P(B\cap A^{\prime })=0.69-0.35=0.34$
Hence $P(B\cap A^{\prime })=0.34$