JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Probability XII

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A student appears for a quiz consisting of only true-false type questions and answers all the questions. The student knows the answers of some questions and guesses the answers for the remaining questions. Whenever the student knows the answer of a question, he gives the correct answer. Assume that the probability of the student giving the correct answer for a question, given that he has guessed it, is

\frac{1}{2}

. Also assume that the probability of the answer for a question being guessed, given that the student's answer is correct, is

\frac{1}{6}

. Then the probability that the student knows the answer of a randomly chosen question is :

(A) $\frac{1}{12}$
(B) $\frac{1}{7}$
(C) $\frac{5}{7}$ ✓ Correct
(D) $\frac{5}{12}$

Answer: (C)

Solution

Let

K

= event that student knows the answer,

G

= event that student guesses,

C

= event that answer is correct.

Given:

P (C ∣ G) = \frac{1}{2}

P (C ∣ K) = 1

P (G ∣ C) = \frac{1}{6}

.

Let

P (K) = x

, then

P (G) = 1 - x

.

By Bayes' theorem:

P (G ∣ C) = \frac{P ( G ) \cdot P ( C ∣ G )}{P ( G ) \cdot P ( C ∣ G ) + P ( K ) \cdot P ( C ∣ K )}

\frac{1}{6} = \frac{( 1 - x ) \cdot \frac{1}{2}}{( 1 - x ) \cdot \frac{1}{2} + x \cdot 1}

\frac{1}{6} = \frac{\frac{1 - x}{2}}{\frac{1 - x}{2} + x}

Cross-multiplying and solving:

\frac{1 - x}{2} + x = 6 \cdot \frac{1 - x}{2}

1 - x + 2 x = 6 (1 - x)

1 + x = 6 - 6 x

7 x = 5 \Rightarrow x = \frac{5}{7}

JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Single correct option

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JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Probability XII

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