25% of the population are smokers. A smoker has 27 times more chances to develop lung cancer than a non-smoker. A person is diagnosed with lung cancer and the probability that this person is a smoker is  $\frac{k}{10}$ then the value of k=___

Let events
$E_1$ : Selected person smoker 
$E_2$ : Selected person  non smoker
E: Selected person is diagnosed with lung cancer.
$\therefore P\left(E_1\right)=\frac{1}{4},P\left(E_2\right)=\frac{3}{4},P(E/E_1)=\frac{27}{28}andP(E/E_2)=\frac{1}{28}$

$\Rightarrow P(E/E_1)=\frac{P\left(E_1\right)P(E/E_1)}{p\left(E\right)}$

$=\frac{\frac{1}{4}\times \frac{27}{28}}{\frac{1}{4}\times \frac{27}{28}+\frac{3}{4}\times \frac{1}{28}}=\frac{27}{90}=\frac{9}{10}=\frac{k}{10}\left(given\right)$

$\therefore k=9$