JEE Main 2023 — Mathematics Probability & Statistics

Given,

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

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

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

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

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

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

So the probability of him being smoker (using bayes' theorem) will be $P(\frac{\text{smoker}}{\text{Lung cancer}})=\frac{0.25\times 27p}{0.25\times 27p+0.75\times p}=\frac{27}{30}=\frac{9}{10}$P(Lung cancersmoker​)=0.25×27p+0.75×p0.25×27p​=3027​=109​

Now comparing with $\frac{k}{10}=\frac{9}{10}$10k​=109​,

Then the value of $k$k will be $9$9.