Answer
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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. $0 \leqslant P\left( E \right) \leqslant 1$.
$P\left( {A' \cap B'} \right)$ 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. $0 \leqslant P\left( E \right) \leqslant 1$.
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