Question

In an entrance test, there are multiple choice questions. There are four possible answers to each question, only one of which 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 the probability that he was guessing is

• A. \(\frac{37}{40}\)
• B. \(\frac{1}{37}\)
• C. \(\frac{36}{37}\)
• D. \(\frac{1}{9}\)

Hint:

Use Bayes' theorem. A guess is correct with probability 1/4, and a known answer is always correct.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We want the probability that the student guessed, given that the answer is correct. This is a conditional probability that needs Bayes' theorem.

Step 2: Key Formula or Approach:
Let \(K\) be the event that the student knows the answer and \(G\) the event that the student guesses. Let \(C\) be the event of a correct answer.
\[ P(K) = 0.9,\quad P(G) = 0.1,\quad P(C|K) = 1,\quad P(C|G) = \tfrac14 \]
\[ P(G|C) = \frac{P(G)P(C|G)}{P(K)P(C|K) + P(G)P(C|G)} \]

Step 3: Detailed Explanation:
Numerator: \(0.1\times\tfrac14 = 0.025\).
Denominator: \(0.9\times1 + 0.025 = 0.925\).
\[ P(G|C) = \frac{0.025}{0.925} = \frac{25}{925} = \frac{1}{37} \]
Option (C) \(\tfrac{36}{37}\) is the probability that the student knew the answer given it is correct. Option (A) and (D) do not match the ratio 25/925.

Final Answer:
The probability that the student was guessing is \(\dfrac1{37}\), option (B). \[ \boxed{\frac{1}{37} \text{ (B)}} \]