$A$ and $B$ are two events such that $P\left( A \right)=0.54,P\left( B \right)=0.69\text{ and }P\left( A\cap B \right)=0.35$, now find the value of the following:
A. $P\left( A'\cap B' \right)$
B. $P\left( B\cap A' \right)$

Hint:To solve these types of questions and we will apply the De morgan’s law for the first part that is $\left( A\cup B \right)'=A'\cap B'$ and again apply the formula for expanding the union of set that is $\left( A\cup B \right)=A+B-\left( A\cap B \right)$. Then for the second part, we will apply the set property that is : $\left( B\cap A' \right)=B-\left( A\cap B \right)$ and then we will put the values from the question and get the answer.

Complete step by step answer:
We are given that: $P\left( A \right)=0.54,P\left( B \right)=0.69\text{ and }P\left( A\cap B \right)=0.35$ , now for the first part that is : $P\left( A'\cap B' \right)$.
Now, according the Demorgan’s law that is the complement of the union of two sets $A\text{ and }B$ is equal to the intersection of the complement of the sets $A\text{ and }B$ : $\left( A\cup B \right)'=A'\cap B'$ , where $A'=\left\{ x:x\in \bigcup \text{ and }x\notin A \right\}$ , where $A'$ denotes the complement.
Now, we have $P\left( A'\cap B' \right)$, so by Demorgan law, we have:
\[\begin{align}
  & \Rightarrow P\left( A'\cap B' \right)=P\left( A\cup B \right)' \\
 & \Rightarrow P\left( A'\cap B' \right)=1-P\left( A\cup B \right) \\
\end{align}\]
Now, we know that the union of set is equal to the following: $\left( A\cup B \right)=A+B-\left( A\cap B \right)$ , therefore:
\[\Rightarrow P\left( A'\cap B' \right)=1-P\left( A\cup B \right)=1-\left( P\left( A \right)+P\left( B \right)-P\left( A\cap B \right) \right)\text{ }.......\text{Equation 1}\] ,
We have, $P\left( A \right)=0.54,P\left( B \right)=0.69\text{ and }P\left( A\cap B \right)=0.35$, we will now put these values in equation 1 :
\[\begin{align}
  & \Rightarrow P\left( A'\cap B' \right)=1-P\left( A\cup B \right)=1-\left( P\left( A \right)+P\left( B \right)-P\left( A\cap B \right) \right)\text{ } \\
 & \Rightarrow P\left( A'\cap B' \right)=1-P\left( A\cup B \right)=1-\left( 0.54+0.69-0.35 \right) \\
 & \Rightarrow P\left( A'\cap B' \right)=1-P\left( A\cup B \right)=1-0.88 \\
 & \Rightarrow P\left( A'\cap B' \right)=0.12 \\
\end{align}\]
Now, we will consider the second part of the question that is: $P\left( B\cap A' \right)$ ,
Now, we know that: $\left( B\cap A' \right)=B-\left( A\cap B \right)$ ,
So, therefore: $P\left( B\cap A' \right)=P\left( B \right)-P\left( A\cap B \right)\text{ }........\text{Equation 2}$
We have, $P\left( B \right)=0.69\text{ and }P\left( A\cap B \right)=0.35$, we will now put these values in equation 2:
$\begin{align}
  & \Rightarrow P\left( B\cap A' \right)=P\left( B \right)-P\left( A\cap B \right) \\
 & \Rightarrow P\left( B\cap A' \right)=0.69-0.35 \\
 & \Rightarrow P\left( B\cap A' \right)=0.34 \\
\end{align}$

Hence, the answer is:
$\begin{align}
  & \Rightarrow P\left( A'\cap B' \right)=0.12 \\
 & \Rightarrow P\left( B\cap A' \right)=0.34 \\
\end{align}$

Note:
 Although the calculations are easy in these types of questions students can make mistakes while applying the formulas. For example, let’s say a student writes the formula wrong for\[P\left( A'\cap B' \right)\] as $P\left( A'\cap B' \right)=P\left( A' \right)\cap P\left( B' \right)$, this will give us: $P\left( A'\cap B' \right)=0.54-0.69=-0.15$ , which can’t be true as probability can’t be negative. So, one must know the formulas from the set theory. De Morgan laws can also be applied to other mathematical topics as well not only in probability theory.