Q65 / 180 · KCET 2019 · Mathematics

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

  • A
    \( \frac{1}{5} \)
  • Correct answer
    B
    \( \frac{3}{5} \)
  • C
    \( \frac{2}{5} \)
  • D
    \( \frac{1}{10} \)

Solution

$\mathrm{S}=\{1,2,3,4,5,6,7\}$ $E_{1}=$ An even number is picked, $E_{2}=$ An odd number is picked $P\left(E_{1}\right)=\frac{3}{7}, P\left(E_{2}\right)=\frac{4}{7}$ $E: A$ man reports an even number $P\left(E \mid E_{1}\right)=\frac{2}{3}$ $P\left(E \mid E_{2}\right)=\frac{1}{3}$ Required probability $=P\left(E_{1} \mid E\right)$ $=\frac{P\left(E \mid E_{1}\right) P\left(E_{1}\right)}{P\left(E \mid E_{1}\right) P\left(E_{1}\right)+P\left(E \mid E_{2}\right) P\left(E_{2}\right)}$ $=\frac{\left(\frac{2}{3}\right)\left(\frac{3}{7}\right)}{\left(\frac{2}{3}\right)\left(\frac{3}{7}\right)+\left(\frac{1}{3}\right)\left(\frac{4}{7}\right)}=\frac{6}{6+4}=\frac{3}{5}$

Hence, the correct answer is option: 'B'.

Relevant Questions