JEE Advanced 1991

LEVELJEE Main

Animated Solution for Mathematics - Probability: In a test an examine either guesses or copies or knows the answer to a multiple choice question with four choices. The probability that he make a guess is $1/3$ and the probability that he copies the answer is $1/6$ . The probability that his answer is correct given that he copied it, is $1/8$ . Find the probability that he knew the answer to the question given that he correctly answered it.

Progressive Hints

Hint 1

The Spark: Visualize the student's journey as a tree diagram with three branches: Guessing, Copying, and Knowing. These branches represent the mutually exclusive and exhaustive ways the student arrives at an answer. The total probability of getting the answer correct is the sum of the probabilities of these three paths.

Hint 2

The Arsenal: You are looking for the probability of a 'cause' (knowing the answer) given an 'effect' (the answer is correct). This is the classic domain of Bayes' Theorem:

P (A ∣ B) = \frac{P ( B ∣ A ) P ( A )}{P ( B )}

Hint 3

The Execution: Start by calculating the total probability of the correct answer,

P (C)

, by summing the probabilities of all three paths:

P (C) = P (A_{1}) P (C ∣ A_{1}) + P (A_{2}) P (C ∣ A_{2}) + P (A_{3}) P (C ∣ A_{3})

. Then, use this to find the conditional probability

P (A_{3} ∣ C) = \frac{P ( A _{3} ) P ( C ∣ A _{3} )}{P ( C )}

Visualized Solution

Defining the Base Events

Let $A_{1}$ be the event that the examinee guesses the answer.
Let $A_{2}$ be the event that the examinee copies the answer.
Let $A_{3}$ be the event that the examinee knows the answer.

Calculating $P (A_{3})$

Given: $P (A_{1}) = \frac{1}{3}$ and $P (A_{2}) = \frac{1}{6}$ .
Since $A_{1}, A_{2}, A_{3}$ are mutually exclusive and exhaustive: $P (A_{1}) + P (A_{2}) + P (A_{3}) = 1$ .
$P (A_{3}) = 1 - (\frac{1}{3} + \frac{1}{6}) = 1 - \frac{1}{2} = \frac{1}{2}$ .

The Event of Answering Correctly

Let $C$ be the event that the examinee's answer is correct.
We need to find the probability of getting the correct answer under each scenario.

Conditional Probabilities for Correct Answer

$P (C ∣ A_{1}) = \frac{1}{4}$ (Probability of a correct guess in a 4-choice MCQ).
$P (C ∣ A_{2}) = \frac{1}{8}$ (Given probability of being correct when copying).
$P (C ∣ A_{3}) = 1$ (If he knows the answer, it is definitely correct).

Identifying the Required Probability

We are given that the answer is correct (Event $C$ has occurred).
We need to find the probability that he knew the answer: $P (A_{3} ∣ C)$ .
This is a classic reverse probability problem, requiring Bayes' Theorem.

Applying Bayes' Theorem

$P (A_{3} ∣ C) = \frac{P ( A _{3} ) P ( C ∣ A _{3} )}{P ( A _{1} ) P ( C ∣ A _{1} ) + P ( A _{2} ) P ( C ∣ A _{2} ) + P ( A _{3} ) P ( C ∣ A _{3} )}$
The numerator is the favorable path (Knowing and Correct).
The denominator is the Total Probability of being correct.

Substituting the Values

Numerator: $\frac{1}{2} \times 1$
Denominator: $(\frac{1}{3} \times \frac{1}{4}) + (\frac{1}{6} \times \frac{1}{8}) + (\frac{1}{2} \times 1)$
$P (A_{3} ∣ C) = \frac{\frac{1}{2}}{\frac{1}{12} + \frac{1}{48} + \frac{1}{2}}$

Evaluating the Denominator

Total Probability $P (C) = \frac{1}{12} + \frac{1}{48} + \frac{1}{2}$
Taking LCM as $48$ :
$P (C) = \frac{4}{48} + \frac{1}{48} + \frac{24}{48} = \frac{29}{48}$

Final Probability

$P (A_{3} ∣ C) = \frac{\frac{1}{2}}{\frac{29}{48}}$
$P (A_{3} ∣ C) = \frac{1}{2} \times \frac{48}{29}$
$P (A_{3} ∣ C) = \frac{24}{29}$

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The Sigma Insight: Bayes' Theorem

Solution Diagram

Analyzing the Setup

To solve this problem, we define the sample space using three mutually exclusive and exhaustive events:

A_{1}

: The student guesses the answer.

A_{2}

: The student copies the answer.

A_{3}

: The student knows the answer.

We are given the prior probabilities as

P (A_{1}) = \frac{1}{3}

and

P (A_{2}) = \frac{1}{6}

. Since the sum of all probabilities must equal

1

, we calculate the probability of the student knowing the answer as:

P (A_{3}) = 1 - (\frac{1}{3} + \frac{1}{6}) = \frac{1}{2}

Defining Conditional Probabilities

Let

C

be the event that the student answers the question correctly. Based on the problem constraints, we define the conditional probabilities of getting the answer correct given each path:

P (C ∣ A_{1}) = \frac{1}{4}

(Probability of guessing correctly)

P (C ∣ A_{2}) = \frac{1}{8}

(Probability of copying correctly)

P (C ∣ A_{3}) = 1

(Probability of knowing correctly)

The Master Equation

We seek the posterior probability

P (A_{3} ∣ C)

, which represents the probability that the student knew the answer given that they answered correctly. According to Bayes' Theorem:

P (A_{3} ∣ C) = \frac{P ( A _{3} ) P ( C ∣ A _{3} )}{P ( C )}

The numerator is the probability of the 'knowing' path:

P (A_{3}) \times P (C ∣ A_{3}) = \frac{1}{2} \times 1 = \frac{1}{2}

The denominator is the total probability of the event

C

, calculated using the Law of Total Probability:

P (C) = P (A_{1}) P (C ∣ A_{1}) + P (A_{2}) P (C ∣ A_{2}) + P (A_{3}) P (C ∣ A_{3})

Final Calculation

Substituting the known values into the total probability equation:

P (C) = (\frac{1}{3} \times \frac{1}{4}) + (\frac{1}{6} \times \frac{1}{8}) + (\frac{1}{2} \times 1)

P (C) = \frac{1}{12} + \frac{1}{48} + \frac{1}{2} = \frac{4 + 1 + 24}{48} = \frac{29}{48}

Finally, we compute the posterior probability:

P (A_{3} ∣ C) = \frac{1/2}{29/48} = \frac{24}{29}

The probability that the student actually knew the answer is $\frac{24}{29}$ .

Common Traps & FAQs

Why is

P (C ∣ A_{3}) = 1

If the student knows the answer, the probability of getting it correct is absolute certainty. There is no randomness involved when knowledge is present, so the conditional probability is

1

Why can't we just use simple probability?

Simple probability works for forward events. Here, we are observing the outcome (the correct answer) and working backward to infer the cause (the student knew it). This 'inverse' nature is exactly why Bayes' Theorem is required.

What if the sum of

P (A_{1}), P (A_{2}), P (A_{3})

was not 1?

In this problem, the events are exhaustive, meaning they cover all possibilities. If they didn't sum to 1, it would imply there's another way the student could answer, which would change the entire sample space. Always ensure your events are exhaustive before applying the Law of Total Probability.

Visualized Solution

Defining the Base Events

Calculating $P (A_{3})$

The Event of Answering Correctly

Conditional Probabilities for Correct Answer

Identifying the Required Probability

Applying Bayes' Theorem

Substituting the Values

Evaluating the Denominator

Final Probability

The Sigma Insight: Bayes' Theorem

Analyzing the Setup

Defining Conditional Probabilities

The Master Equation

Final Calculation

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Visualized Solution

Defining the Base Events

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(A3​)

The Event of Answering Correctly

Conditional Probabilities for Correct Answer

Identifying the Required Probability

Applying Bayes' Theorem

Substituting the Values

Evaluating the Denominator

Final Probability

The Sigma Insight: Bayes' Theorem

Analyzing the Setup

Defining Conditional Probabilities

The Master Equation

Final Calculation

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