JEE Main 2023 (25 January Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Probability: 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 $k$ is

Q: What if the question asked for the probability that a non-smoker has cancer given they are diagnosed? How would the formula change?

You would simply calculate $P(E_2|E)$. The numerator would become $P(E_2)P(E|E_2)$, while the denominator (the total probability of having cancer) remains exactly the same. It is the same 'pie' being sliced differently!

Enter Numerical Value:

Visualized Solution

Defining the Population Events

Let $E_{1}$ be the event that a person is a smoker.
Let $E_{2}$ be the event that a person is a non-smoker.

Calculating Prior Probabilities

$P (E_{1}) = \frac{25}{100} = \frac{1}{4}$
$P (E_{2}) = \frac{75}{100} = \frac{3}{4}$

Defining the Condition: Lung Cancer

Let $E$ be the event that the person is diagnosed with lung cancer.

Conditional Probabilities

Let $P (E ∣ E_{2}) = p$ (Probability for non-smoker)
Then $P (E ∣ E_{1}) = 27 p$ (Probability for smoker)

The Goal and Bayes' Theorem

We need to find $P (E_{1} ∣ E)$ .
By Bayes' Theorem:
$P (E_{1} ∣ E) = \frac{P ( E _{1} ) P ( E ∣ E _{1} )}{P ( E _{1} ) P ( E ∣ E _{1} ) + P ( E _{2} ) P ( E ∣ E _{2} )}$

Substituting the Values

Substitute the values into the formula:
$P (E_{1} ∣ E) = \frac{\frac{1}{4} \times 27 p}{\frac{1}{4} \times 27 p + \frac{3}{4} \times p}$

Simplifying the Expression

Cancel out $p$ and the common factor $\frac{1}{4}$ :
$P (E_{1} ∣ E) = \frac{27}{27 + 3}$

Final Calculation

Simplify the denominator:
$P (E_{1} ∣ E) = \frac{27}{30}$
$P (E_{1} ∣ E) = \frac{9}{10}$

Finding the value of k

Given $P (E_{1} ∣ E) = \frac{k}{10}$
Comparing both sides:
$\frac{9}{10} = \frac{k}{10} ⟹ k = 9$

00:00 / 00:00

The Sigma Insight: Bayes' Theorem

Solution Diagram

Analyzing the Setup

To solve this problem, we define our sample space using two mutually exclusive events:

E_{1}

: The patient is a smoker.

E_{2}

: The patient is a non-smoker.

Given that 25% of the population are smokers, we have:

P (E_{1}) = \frac{25}{100} = \frac{1}{4}

Since the groups are mutually exclusive and exhaustive, the probability of being a non-smoker is:

P (E_{2}) = 1 - P (E_{1}) = \frac{3}{4}

The Weight of Evidence

Let

E

be the event that the patient is diagnosed with lung cancer. We are given that a smoker is 27 times more likely to develop lung cancer than a non-smoker.

If we define the probability of a non-smoker developing cancer as

p

, we have:

P (E ∣ E_{2}) = p

Consequently, the probability of a smoker developing cancer is:

P (E ∣ E_{1}) = 27 p

The Bayesian Bridge

We seek the posterior probability

P (E_{1} ∣ E)

, which represents the probability that the patient is a smoker given they have been diagnosed with lung cancer. According to Bayes' Theorem:

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 our known values into the equation, we obtain:

P (E_{1} ∣ E) = \frac{\frac{1}{4} \times 27 p}{\frac{1}{4} \times 27 p + \frac{3}{4} \times p}

The Elegance of Cancellation

We observe that the variable

p

is present in every term of the numerator and the denominator. We can safely divide both by

p

to eliminate it.

Furthermore, we can multiply the numerator and the denominator by 4 to clear the fractions:

P (E_{1} ∣ E) = \frac{27}{27 + 3}

Simplifying the expression:

P (E_{1} ∣ E) = \frac{27}{30} = \frac{9}{10}

Final Calculation

The problem states that the resulting probability is equal to

\frac{k}{10}

. By equating our result to this form:

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

Thus, the value of $k = 9$ .

Common Traps & FAQs

Wait, why do we just assume

P (E ∣ E_{2}) = p

? Don't we need the actual probability of getting cancer?

That is a brilliant observation! In many JEE problems, you are given relative ratios rather than absolute values. Because we are calculating a conditional probability (a ratio of probabilities), the absolute value of

p

acts as a common factor in both the numerator and the denominator, allowing it to cancel out entirely. You don't need the absolute risk to find the relative likelihood.

Why can't I just say the probability is 27/28 since the smoker is 27 times more likely?

That is a common trap! You are forgetting to weight the probabilities by the size of the population groups. A smoker is 27 times more likely to get cancer, but there are three times as many non-smokers in the population. You must multiply the individual risk by the group's prevalence (the prior probability) before comparing them.

Is Bayes' Theorem always this straightforward, or am I missing a hidden complexity?

Bayes' Theorem is essentially a way of 'reversing' conditional probability. As long as you clearly define your events

E_{1}

and

E_{2}

(the causes) and

E

(the effect), the math is just a systematic accounting of the total probability space. The complexity in JEE usually lies in correctly identifying the prior probabilities, not the formula itself.

What if the question asked for the probability that a non-smoker has cancer given they are diagnosed? How would the formula change?

You would simply calculate

P (E_{2} ∣ E)

. The numerator would become

P (E_{2}) P (E ∣ E_{2})

, while the denominator (the total probability of having cancer) remains exactly the same. It is the same 'pie' being sliced differently!

Why does the

1/4

factor cancel out? Is that always allowed?

Yes, it is allowed because it appears in every term of the numerator and denominator. Mathematically, you are factoring out

1/4

from the entire fraction:

\frac{( 1/4 ) \times X}{( 1/4 ) \times X + ( 1/4 ) \times Y} = \frac{1/4 ( X )}{1/4 ( X + Y )}

. Since

1/4

is not zero, it cancels out perfectly.

Animated Solution for Mathematics - Probability: 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 $k$ is

Enter Numerical Value:

Visualized Solution

Defining the Population Events

Calculating Prior Probabilities

Defining the Condition: Lung Cancer

Conditional Probabilities

The Goal and Bayes' Theorem

Substituting the Values

Simplifying the Expression

Final Calculation

Finding the value of k

The Sigma Insight: Bayes' Theorem

Analyzing the Setup

The Weight of Evidence

The Bayesian Bridge

The Elegance of Cancellation

Final Calculation

Similar Questions

A bag contains 10 balls out of which $k$ are red and $(10 - k)$ are black, where $0 \leq k \leq 10$ . If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is:

A card from a pack of 52 cards is lost. From the remaining 51 cards, $n$ cards are drawn and are found to be spades. If the probability of the lost card to be a spade is $\frac{11}{50}$ , then $n$ is equal to