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 which occurs together with any one of E1 or E2 or … or En such that P(A) ≠ 0.

Then \(\rm P\left( {{E_i}\;|\;A} \right) = \frac{{P\left( {{E_i}} \right)\; \times \;P\left( {A\;|\;{E_i}} \right)}}{{\mathop \sum \nolimits_{i\; = 1}^n P\left( {{E_i}} \right) \times P\left( {A\;|\;{E_i}} \right)}},\;i = 1,\;2,\; \ldots .\;,\;n\)

Calculation:

Let E₁: He knows the answer

E2: He does not know the answer

X: He gets the correct answer.

Therefore, P (E1) = 90% = \(\frac {9}{10}\)

P (E2) = \(1- \frac {9}{10} = \frac {1}{10}\) 

P (X | E1)  = 1

P (X | E2)  = \(\frac 1 4\)

As we know that according to Bayes' theorem:

 \(\rm P\left( {{E_i}\;|\;A} \right) = \frac{{P\left( {{E_i}} \right)\; \times \;P\left( {A\;|\;{E_i}} \right)}}{{\mathop \sum \nolimits_{i\; = 1}^n P\left( {{E_i}} \right) \times P\left( {A\;|\;{E_i}} \right)}},\;i = 1,\;2,\; \ldots .\;,\;n\)

\(\therefore \rm P{\rm{(}}{E_2}\;{\rm{|}}X) = \frac{{\frac{1}{10} \cdot \frac{1}{{4}}}}{{\left[ {\frac{9}{10} \cdot 1 + \;\frac{1}{10} \cdot \frac{1}{{4}}} \right]}} = \frac{{1}}{{37}}\;\)