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Question

In an entrance test there are multiple choice questions. There are four options for each question, of which only one is correct. The probability that a student knows the answer to a question is 90%. If he gets the correct answer to a question, then what is the probability that he was guessing ?

This question was previously asked in
NDA 1 2026 GAT Question Paper (12-Apr-2026)
The correct answer is
1/37

Problem Setup: Probability Events

We need to calculate the conditional probability that a student guessed the answer, given they got it right. This involves understanding conditional probability and applying Bayes' Theorem.

  • Event K: Student knows the answer. The probability is given as \(P(K) = 0.90\).
  • Event G: Student guesses the answer. This is the complement of knowing the answer, so \(P(G) = 1 - P(K) = 1 - 0.90 = 0.10\).
  • Event C: Student answers the question correctly.
  • If the student knows the answer, the probability of answering correctly is certain: \(P(C | K) = 1\).
  • If the student guesses, there are four options, so the probability of guessing correctly is \(P(C | G) = \frac{1}{4}\).

Applying Bayes' Theorem Formula

The question asks for the probability that the student was guessing given that they answered correctly. This is represented as \(P(G | C)\).

Bayes' Theorem provides the formula:

\(P(G | C) = \frac{P(C | G) \times P(G)}{P(C)}\)

Calculating Overall Probability of Correct Answer

To use Bayes' Theorem, we first need to find the total probability of answering correctly, \(P(C)\). We can calculate this using the law of total probability, considering both cases: knowing the answer and guessing.

\(P(C) = P(C | K) \times P(K) + P(C | G) \times P(G)\)

Substitute the known values:

\(P(C) = (1 \times 0.90) + (\frac{1}{4} \times 0.10)\)

\(P(C) = 0.90 + 0.025\)

\(P(C) = 0.925\)

Final Probability Calculation: Guessing Given Correct Answer

Now, substitute the calculated \(P(C)\) and the given probabilities into Bayes' Theorem to find \(P(G | C)\):

\(P(G | C) = \frac{P(C | G) \times P(G)}{P(C)}\)

\(P(G | C) = \frac{0.025}{0.925}\)

To simplify this fraction, we can convert the decimals to fractions:

\(0.025 = \frac{25}{1000} = \frac{1}{40}\)

\(0.925 = \frac{925}{1000} = \frac{37}{40}\)

Now perform the division:

\(P(G | C) = \frac{1/40}{37/40}\)

\(P(G | C) = \frac{1}{40} \times \frac{40}{37}\)

\(P(G | C) = \frac{1}{37}\)

The probability that the student was guessing, given they answered correctly, is 1/37.

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Important Questions from Bayes' Theorem

  1. A certain disease is difficult to be diagnosed and the probability of correctly diagnosing the disease is 0.6. If any patient, after the correct diagnosis, has 40% chances of dying. However, an incorrect diagnosis enhances the probability of death to 0.7. If a patient has died after the treatment, what is the probability that the disease was diagnosed correctly?

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