Question

Probability that $A$ speaks truth is $\frac{4}{5}$. A coin is tossed. A reports that a head appears. The probability that actually there was head is

• A. $\frac{4}{5}$
• B. $\frac{1}{2}$
• C. $\frac{1}{5}$
• D. $\frac{2}{5}$

Answer 1

The correct option is A $\frac{4}{5}$

Let $E_1$ and $E_2$ be the events such that
$E_1:A$ speaks truth
$E_2:A$ speaks false
Let $X$ be the event that a head appears.
$P\left(E_1\right)=\frac{4}{5}$
$\therefore P\left(E_2\right)=1-P\left(E_1\right)=1-\frac{4}{5}=\frac{1}{5}$
If a coin is tossed, then it may result in either head $\left(H\right)$ or tail $\left(T\right)$.
The probability of getting a head is $\frac{1}{2}$ whether $A$ speaks truth or not.
$\therefore P\left(X|E_1\right)=P\left(X|E_2\right)=\frac{1}{2}$
The probability that there is actually a head is given by $P\left(E_1|X\right)$.
$P\left(E_1|X\right)=\frac{P\left(E_1\right)\cdot P\left(X|E_1\right)}{P\left(E_1\right)\cdot P\left(X|E_1\right)+P\left(E_2\right)\cdot P\left(X|E_2\right)}$
$=\frac{\frac{4}{5}\cdot \frac{1}{2}}{\frac{4}{5}\cdot \frac{1}{2}+\frac{1}{5}\cdot \frac{1}{2}}$
$=\frac{\frac{1}{2}\cdot \frac{4}{5}}{\frac{1}{2}\left(\frac{4}{5}+\frac{1}{5}\right)}$
$=\frac{\frac{4}{5}}{1}$
$=\frac{4}{5}=0.80$