Question

The probability for a contractor to get a road contract is $ \dfrac{2}{3} $ and to get a building contract is $ \dfrac{5}{9} $ . The probability to get at least one contract is $ \dfrac{4}{5}. $ Find the probability that he gets both the contracts.

Answer 1

Hint: If $ A $ and $ B $ are any two events and are not disjoint, then
 $ P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right) $ ._ _ _ _ _ _ _ _ _ _ $ \left( 1 \right) $
 $ A $ and $ B $ are the subsets of sample space S.
This is the additional theorem of probability.

Complete step-by-step answer:
As we know the addition theorem,
 $ P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right) $
Let, $ P\left( A \right) $ be the probability of the contractor to get a road contract $ = \dfrac{2}{3}. $
Let, $ P\left( B \right) $ be the probability of the contractor to get a building contract $ = \dfrac{5}{9}. $
Let, $ P\left( {A \cup B} \right) $ be the probability of the contractor to get at least one contract $ = \dfrac{4}{5}. $
Let, $ P\left( {A \cap B} \right) $ be the probability of contractor to get both contract $ = ? $
Now, using equation $ \left( 1 \right) $ ,
 $ \Rightarrow $ $ P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right) $
 $ \Rightarrow \dfrac{4}{5} = \dfrac{3}{2} + \dfrac{5}{9} - P\left( {A \cap B} \right) $ $ $
 $
   \Rightarrow P\left( {A \cap B} \right) = \dfrac{2}{3} + \dfrac{5}{9} - \dfrac{4}{5} \\
   \Rightarrow P\left( {A \cap B} \right) = \dfrac{{45 \times 2 + 5 \times 15 - 4 \times 27}}{{135}} \\
   \Rightarrow P\left( {A \cap B} \right) = \dfrac{{90 + 75 - 108}}{{135}} \\
   \Rightarrow P\left( {A \cap B} \right) = \dfrac{{57}}{{135}} \\
   \Rightarrow P\left( {A \cap B} \right) = \dfrac{{19}}{{45}}. \;
  $
Therefore, $ P\left( {A \cap B} \right) $ of contractor to get both contract is $ \dfrac{{19}}{{45}}. $
So, the correct answer is “ $ \dfrac{{19}}{{45}} $ ”.

Note: $ \Rightarrow $ When selecting terms between $ P\left( {A \cap B} \right) $ and $ P\left( {A \cup B} \right) $ , choose wisely, if any terms goes wrong then the whole answer goes wrong.
 $ \Rightarrow $ There is one other important theorem, multiplication theorem,
 $ P\left( {A \cap B} \right) = P\left( A \right).P\left( B \right) $ .