NIMCET 2017 — Mathematics PYQ

Concept:

\textbf{Bayes' Theorem:}

Let $E_1,E_2,\dots ,E_n$ 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 $E_1$ or $E_2$ or $\dots$ or $E_n$ such that $P(A)\ne 0$.

Then $P(E_i\mid A)=\frac{P(E_i)\times P(A\mid E_i)}{{\sum }_{i=1}^{n}P(E_i)\times P(A\mid E_i)},i=1,2,\dots ,n$

Calculation

Let $E_1$: He knows the answer  
$E_2$: He does not know the answer  
$X$: He gets the correct answer  

Therefore, $P(E_1)=90\%=\frac{9}{10}$  

$P(E_2)=1-\frac{9}{10}=\frac{1}{10}$  

$P(X\mid E_1)=1$  

$P(X\mid E_2)=\frac{1}{4}$  

As we know that according to Bayes' theorem:  

$P(E_i\mid A)=\frac{P(E_i)\times P(A\mid E_i)}{{\sum }_{i=1}^{n}P(E_i)\times P(A\mid E_i)},i=1,2,\dots ,n$  

$\therefore P(E_2\mid X)=\frac{\frac{1}{10}\times \frac{1}{4}}{\frac{9}{10}\times 1+\frac{1}{10}\times \frac{1}{4}}=\frac{1}{37}$