Question

A and B are two events such that $P\left( A \right) = 0.54$ , $P\left( B \right) = 0.69$ and$P\left( {A \cap B} \right) = 0.35$. Find  $P\left( {A \cup B} \right)$ $P\left( {A' \cap B'} \right)$ $P\left( {A \cap B'} \right)$ $P\left( {B \cap A'} \right)$

Hint: Here $P\left( {A \cup B} \right)$ means $P\left( {A{\text{ or }}B} \right)$
$P\left( {A' \cap B'} \right)$ means $P\left( {{\text{neither }}A{\text{ nor }}B} \right)$
$P\left( {A \cap B'} \right)$ means $P\left( {A{\text{ but not }}B} \right)$
$P\left( {B \cap A'} \right)$ means $P\left( {B{\text{ but not }}A} \right)$.
Given that,
$P\left( A \right) = 0.54$ , $P\left( B \right) = 0.69$ and $P\left( {A \cap B} \right) = 0.35$

$P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right) \\ {\text{ = }}0.54 + 0.69 - 0.35 \\ {\text{ = 0}}{\text{.88}} \\ \therefore P\left( {A \cup B} \right) = 0.88 \\$

$P\left( {A' \cap B'} \right) = 1 - \left( {P\left( {A \cup B} \right)} \right) \\ {\text{ = }}1 - \left( {0.88} \right) \\ {\text{ = }}0.12 \\ \therefore P\left( {A' \cap B'} \right) = 0.12 \\$
Now consider$P\left( {A \cap B} \right) = P\left( A \right) + P(B) - P\left( {A \cup B} \right) \\ {\text{ = }}0.54 + 0.69 - 0.88 \\ {\text{ = }}0.35 \\ \therefore P\left( {A \cap B} \right) = 0.35 \\$
$P\left( {A \cap B'} \right) = P\left( A \right) - P\left( {A \cap B} \right) \\ {\text{ = }}0.54 - 0.35 \\ {\text{ = }}0.19 \\ \therefore P\left( {A \cap B'} \right) = 0.19 \\$
$P\left( {B \cap A'} \right) = P\left( B \right) - P\left( {A \cap B} \right) \\ {\text{ = }}0.69 - 0.35 \\ {\text{ = }}0.34 \\ \therefore P\left( {B \cap A'} \right) = 0.34 \\$
Thus,
$P\left( {A \cup B} \right) = 0.88 \\ P\left( {A' \cap B'} \right) = 0.12 \\ P\left( {A \cap B'} \right) = 0.19 \\ P\left( {B \cap A'} \right) = 0.34 \\$
Note: Probability of an event always lies between $0$and $1$ i.e. $0 \leqslant P\left( E \right) \leqslant 1$.