Question

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. $\frac{1}{10}$
• B. $\frac{2}{5}$
• C. $\frac{3}{5}$
• D. $\frac{1}{5}$

Answer 1

Answer: (C)

$\frac{3}{5}$

Answer 2

S=\left\\{1,2,3,4,5,6,7\right\\}
$E_{1}=$ 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
$P\left(E| E_{1}\right)=\frac{2}{3}$
$P\left(E |E_{2}\right)=\frac{1}{3}$
$Required \, probability =P\left(E_{1}|E\right)$
$\frac{P\left(E |E_{1}\right)P\left(E_{1}\right)}{P\left(E| E_{1}\right)P\left(E_{1}\right)+P\left(E| E_{2}\right)P\left(E_{2}\right)}$
$=\frac{\left(\frac{2}{3}\right)\left(\frac{3}{7}\right)}{\left(\frac{2}{3}\right)\left(\frac{3}{7}\right)+\left(\frac{1}{3}\right)\left(\frac{4}{7}\right)}=\frac{6}{6+4}=\frac{3}{5}$