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,
$P\left( {A \cap B} \right) = P\left( A \right).P\left( B \right)$ .