In an entrance test there are multiple choice questions, with four possible answer to each question of which one is correct, The probability that a student knows the answer to a question is 90%, If the student gets the correct answer to a question, then the probability that he was guessing is
Concept:
Bayes' Theorem:
Let E1, E2, ..., En be n mutually exclusive and exhaustive events associated with a random experiment, and let S be the sample space.
Let A be any event that occurs together with any one of E1, E2, ..., or En, such that P(A) ≠ 0.
Then, according to Bayes' theorem:
\[\rm P\left( E_i\;|\;A \right) = \frac{P\left( E_i \right) \times P\left( A\;|\;E_i \right)}{\sum\limits_{i = 1}^{n} P\left( E_i \right) \times P\left( A\;|\;E_i \right)},\quad i = 1,2,\dots,n\]
Calculation:
Let:
E1: He knows the answer
E2: He does not know the answer
X: He gets the correct answer.
Given:
P(E1) = 90% = \(\frac{9}{10}\)
P(E2) = \(1 - \frac{9}{10} = \frac{1}{10}\)
P(X | E1) = 1
P(X | E2) = \(\frac{1}{4}\)
Using Bayes' theorem:
\[\rm P(E_2\;|\;X) = \frac{P(E_2) \cdot P(X\;|\;E_2)}{P(E_1) \cdot P(X\;|\;E_1) + P(E_2) \cdot P(X\;|\;E_2)}\]
\[\rm P(E_2\;|\;X) = \frac{\frac{1}{10} \times \frac{1}{4}}{\left( \frac{9}{10} \times 1 + \frac{1}{10} \times \frac{1}{4} \right)} = \frac{1}{37}\]
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