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

$P\left(R\right) = Probability$ of getting road contract $=\frac{2}{3}$
$P\left(B\right)=Probability$ of getting building contract$=\frac{5}{9}$

$P(B\cup R)=\text{ Probability }$ of getting at least one contract$=\frac{4}{5}$

$P(B\cap R)=\text{ Probability }$ of both getting contracts =? 
By addition theorem

$P(B\cup R)=P\left(B\right)+P\left(R\right)-P(B\cap R)$

$\frac{4}{5}=\frac{2}{3}+\frac{5}{9}-P(B\cap R)$

$P(B\cap R)=\frac{2}{3}+\frac{5}{9}-\frac{4}{5}=\frac{30+25-36}{45}$

$=\frac{25-6}{45}=\frac{19}{45}$