Question

Probability that A speaks truth is 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

Let $E_1$ : the event that coin comes up with a head,

$E_2$ : the event that coin comes up with a tail.

Then, $E_1$ and $E_2$ are mutually exclusive and exhaustive events. Moreover,

P$E_1$=P$E_2$= $\frac{1}{2}$

E: The event that A reports that a head appears,

$\therefore P\left(\frac{E}{E_1}\right)$ = P (head comes up and A speaks truth) = $\frac{4}{5}$

$P\left(\frac{E}{E_2}\right)$ = P (head comes up and A speaks truth) = $\frac{4}{5}$

The probability that there is actually a head is given by $P\left(\frac{E_1}{E}\right)$

By using Baye's theorem, we obtain

$P\left(\frac{E}{E_1}\right)=\frac{P\left(\frac{E}{E_1}\right)P\left(E_1\right)}{P\left(\frac{E}{E_1}\right)P\left(E_1\right)+P\left(\frac{E}{E_2}\right)P\left(E_2\right)}=\frac{\frac{4}{5}\times \frac{1}{2}}{\frac{4}{5}\times \frac{1}{2}+\frac{1}{5}\times \frac{1}{2}}=\frac{4}{4+1}=\frac{4}{5}$

Hence, (a) is the correct option.