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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
  $P\left( {A \cup B} \right)$
 $P\left( {A' \cap B'} \right)$
 $P\left( {A \cap B'} \right)$
 $P\left( {B \cap A'} \right)$

Answer Verified Verified
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. $0 \leqslant P\left( E \right) \leqslant 1$.
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