JEE Main 2023 — Probability Question with Solution

Given,

In a city, $25\%$ of the population is smoker

So, $P\left(\text{smoker}\right)=0.25$

$P\left(\text{non-smoking}\right)=0.75$

And smoker has $27$ times more chance of being diagnosed with lung cancer,

So, if $P\left(\frac{\text{Lung cancer}}{\text{non-smoker}}\right)=p$ then $P\left(\frac{\text{Lung cancer}}{\text{smoker}}\right)=27p$,

Now,  a person is selected at random and found to be diagnosed with lung cancer,

So, total probability of having a lung cancer will be 

$P\left(\text{Lung cancer}\right)=0.25\times 27p+0.75\times p$

So the probability of him being smoker (using bayes' theorem) will be $P\left(\frac{\text{smoker}}{\text{Lung cancer}}\right)=\frac{0.25\times 27p}{0.25\times 27p+0.75\times p}=\frac{27}{30}=\frac{9}{10}$

Now comparing with $\frac{k}{10}=\frac{9}{10}$,

Then the value of $k$ will be $9$.