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

The correct option is C $\frac{3}{5}$

$S=\{1,2,3,4,5,6,7\}$
$E_1=$ An even number is picked,
$E_2=$ An odd number is picked
$\therefore P\left(E_1\right)=\frac{3}{7},P\left(E_2\right)=\frac{4}{7}$
$E=$ A man reports that picked number is an even number
$P\left(E|E_1\right)=\frac{2}{3}$ and $P\left(E|E_2\right)=\frac{1}{3}$
Probability that picked number is actually even
$P\left(E_1|E\right)=\frac{P\left(E|E_1\right)P\left(E_1\right)}{P\left(E|E_1\right)P\left(E_1\right)+P\left(E|E_2\right)P\left(E_2\right)}$
$\frac{\left(\frac{2}{3}\right)\left(\frac{3}{7}\right)}{\left(\frac{2}{3}\right)\left(\frac{3}{7}\right)+\left(\frac{1}{3}\right)\left(\frac{4}{7}\right)}=\frac{3}{5}$