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NVAMediumJEE 2023Conditional Probability & Bayes Theorem

JEE Mathematics 2023 Question with Solution

25% of the population are smokers. A smoker has 2727 times more chances to develop lung cancer than a non-smoker. If a person is diagnosed with lung cancer, and the probability that this person is a smoker is k10\frac{k}{10}, then the value of kk is:

Answer

Correct answer:9

Step-by-step solution

Standard Method

Given: Probability that a person is a smoker is P(E1)=14P(E_1)=\frac{1}{4} and probability that a person is a non-smoker is P(E2)=34P(E_2)=\frac{3}{4}.

The event of being diagnosed with lung cancer is EE.

Also, P(E/E1)=2728P(E/E_1)=\frac{27}{28} and P(E/E2)=128P(E/E_2)=\frac{1}{28}.

Find: The value of kk if P(E1/E)=k10P(E_1/E)=\frac{k}{10}.

Using Bayes' theorem,

P(E1/E)=P(E1)P(E/E1)P(E)P(E_1/E)=\frac{P(E_1)P(E/E_1)}{P(E)}

where

P(E)=P(E1)P(E/E1)+P(E2)P(E/E2)P(E)=P(E_1)P(E/E_1)+P(E_2)P(E/E_2)

Substitute the given values:

P(E)=14⋅2728+34⋅128P(E)=\frac{1}{4}\cdot \frac{27}{28}+\frac{3}{4}\cdot \frac{1}{28}

So,

P(E1/E)=14⋅2728P(E)=910P(E_1/E)=\frac{\frac{1}{4}\cdot \frac{27}{28}}{P(E)}=\frac{9}{10}

Thus,

k10=910\frac{k}{10}=\frac{9}{10}

Hence, k=9k=9.

Therefore, the required value is 99.

Expanded Bayes Substitution

Let E1E_1 denote smokers, E2E_2 denote non-smokers, and EE denote diagnosed with lung cancer.

Then,

P(E1)=14,P(E2)=34P(E_1)=\frac{1}{4}, \qquad P(E_2)=\frac{3}{4}

and from the given ratio of chances,

P(E/E1)=2728,P(E/E2)=128P(E/E_1)=\frac{27}{28}, \qquad P(E/E_2)=\frac{1}{28}

Now apply Bayes' theorem:

P(E1/E)=P(E1)P(E/E1)P(E1)P(E/E1)+P(E2)P(E/E2)P(E_1/E)=\frac{P(E_1)P(E/E_1)}{P(E_1)P(E/E_1)+P(E_2)P(E/E_2)}

Substituting,

P(E1/E)=14⋅272814⋅2728+34⋅128=910P(E_1/E)=\frac{\frac{1}{4}\cdot \frac{27}{28}}{\frac{1}{4}\cdot \frac{27}{28}+\frac{3}{4}\cdot \frac{1}{28}}=\frac{9}{10}

Therefore,

k10=910\frac{k}{10}=\frac{9}{10}

which gives k=9k=9.

Common mistakes

  • Interpreting '2727 times more chances' as a direct final probability without normalizing. This is wrong because Bayes' theorem requires conditional probabilities whose total is consistent. Use the ratio 27:127:1 and convert it to 2728\frac{27}{28} and 128\frac{1}{28}.

  • Using P(smoker∣cancer)P(\text{smoker} \mid \text{cancer}) as P(cancer∣smoker)P(\text{cancer} \mid \text{smoker}). These are different conditional probabilities. Apply Bayes' theorem to reverse the conditioning correctly.

  • Ignoring the prior probabilities 14\frac{1}{4} and 34\frac{3}{4}. This is wrong because the population proportion of smokers and non-smokers affects the posterior probability. Always multiply the conditional probability by the corresponding prior.

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