JEE MainMathematics

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Bayes' Theorem Application to Conditional Probability of Disease Risk

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 k/10. Then the value of k is.

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Correct answer

The value of k is 9.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

Let $S$ be the event that a person is a smoker and $N S$ that the person is a non-smoker. $P (S) = 25 % = \frac{1}{4}$ , $P (N S) = 75 % = \frac{3}{4}$ . Let $C$ be the event of developing lung cancer. According to the question, $P (C ∣ S) = 27 P (C ∣ N S)$ .

02Goal

Find the value of $k$ given that the posterior probability $P (S ∣ C) = \frac{k}{10}$ .

03Approach

Use Bayes' Theorem to calculate the posterior probability: $P (S ∣ C) = \frac{P (S) P (C ∣ S)}{P (S) P (C ∣ S) + P (N S) P (C ∣ N S)}$ . Then substitute $P (C ∣ S) = 27 P (C ∣ N S)$ so that the unknown base rate $P (C ∣ N S)$ cancels out.

04Execute

$P (S ∣ C) = \frac{\frac{1}{4} \times 27 P (C ∣ N S)}{\frac{1}{4} \times 27 P (C ∣ N S) + \frac{3}{4} P (C ∣ N S)} = \frac{27}{27 + 3} = \frac{27}{30} = \frac{9}{10}$ Equating $\frac{9}{10} = \frac{k}{10}$ gives $k = 9$ .

✓Verify

Check with population numbers: Out of 100 people, 25 are smokers and 75 are non-smokers. If the cancer rate for non-smokers is $p$ , cancer cases among non-smokers is $75 p$ , and among smokers is $25 \times 27 p = 675 p$ . Total cancer cases $= 750 p$ . Fraction of smokers $= 675 p / 750 p = 675 / 750 = 9 / 10$ . Thus, $k = 9$ .

Hints that build this answer step by step

Let $S$ be the event that a person is a smoker, $S^{'}$ be a non-smoker, and $C$ be the event that a person has lung cancer. If $P (C ∣ S^{'}) = p$ , what are the prior probabilities and conditional probabilities in terms of $p$ ?
$P (S) = \frac{1}{4}$ , $P (S^{'}) = \frac{3}{4}$ , $P (C ∣ S^{'}) = p$ , $P (C ∣ S) = 27 p$
Which formula correctly gives the posterior probability $P (S ∣ C)$ using Bayes' Theorem?
$P (S ∣ C) = \frac{P (S) P (C ∣ S)}{P (S) P (C ∣ S) + P (S^{'}) P (C ∣ S^{'})}$
Substituting the probabilities into the formula, what is the value of $P (S ∣ C)$ and consequently $k$ ?
$P (S ∣ C) = \frac{9}{10}$ , so $k = 9$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

Does '27 times more chances' mean 27 times as likely or (1 + 27) times?

In standard Indian competitive exams like JEE, 'n times more' often means 'n times as much'. Therefore, the likelihood ratio is taken directly as n. Here, that value is 27.

Answer

The value of k is 9.

Why each option works or fails

Step-by-step solution

Given: Let $S$ be the event that a person is a smoker and $N S$ that the person is a non-smoker. $P (S) = 25 % = \frac{1}{4}$ , $P (N S) = 75 % = \frac{3}{4}$ . Let $C$ be the event of developing lung cancer. According to the question, $P (C ∣ S) = 27 P (C ∣ N S)$ .
Goal: Find the value of $k$ given that the posterior probability $P (S ∣ C) = \frac{k}{10}$ .
Approach: Use Bayes' Theorem to calculate the posterior probability: $P (S ∣ C) = \frac{P (S) P (C ∣ S)}{P (S) P (C ∣ S) + P (N S) P (C ∣ N S)}$ . Then substitute $P (C ∣ S) = 27 P (C ∣ N S)$ so that the unknown base rate $P (C ∣ N S)$ cancels out.
Execute: $P (S ∣ C) = \frac{\frac{1}{4} \times 27 P (C ∣ N S)}{\frac{1}{4} \times 27 P (C ∣ N S) + \frac{3}{4} P (C ∣ N S)} = \frac{27}{27 + 3} = \frac{27}{30} = \frac{9}{10}$ Equating $\frac{9}{10} = \frac{k}{10}$ gives $k = 9$ .
Verify: Check with population numbers: Out of 100 people, 25 are smokers and 75 are non-smokers. If the cancer rate for non-smokers is $p$ , cancer cases among non-smokers is $P (N S) = 75 % = \frac{3}{4}$ , and among smokers is $25 \times 27 p = 675 p$ . Total cancer cases $= 750 p$ . Fraction of smokers $= 675 p / 750 p = 675 / 750 = 9 / 10$ . Thus, $k = 9$ .

Bayes' Theorem Application to Conditional Probability of Disease Risk

What feels right?

The value of k is 9.

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

✓Verify

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