UPSC NDA Math Model Paper 1

Given, a man speaks truth 2 out of 3 times. He picks one of the natural numbers in the set S = {1, 2, 3, 4, 5, 6, 7} and reports that it is even. $E_{1}=$ the event of an even number is picked $E_{2}=$ the event of an odd number is picked $\Rightarrow P(E_{1})=\frac{3}{7}$ $\Rightarrow P(E_{2})=\frac{4}{7}$ $E=$ the event a man reports that it is even A man speaks truth 2 out of 3 times.$\Rightarrow P(E \mid E_{1})=\frac{2}{3}$ $\Rightarrow P(E \mid E_{2})=\frac{1}{3}$ To find: The probability that it is actually even i.e. $P(E_{1} \mid E)$ $P(E_{1} \mid E)$ $= \frac{ P(E \mid E_{1}) \cdot P(E_{1}) }{ P(E \mid E_{1}) \cdot P(E_{1}) + P(E \mid E_{2}) \cdot P(E_{2}) }$ $\Rightarrow P(E_{1} \mid E)$ $= \frac{ \left( \frac{2}{3} \right) \cdot \left( \frac{5}{7} \right) }{ \left( \frac{2}{3} \right) \cdot \left( \frac{3}{7} \right) + \left( \frac{1}{3} \right) \cdot \left( \frac{4}{7} \right) }$ $\Rightarrow P(E_{1} \mid E) = \frac{6}{10}$ $\Rightarrow P(E_{1} \mid E) = \frac{3}{5}$ Hence, a man speaks truth 2 out of 3 times. He picks one of the natural numbers in the set S = {1, 2, 3, 4, 5, 6, 7} and reports that it is even. The probability that it is actually even is $\frac{3}{5}$