Probability that A speaks truth is

Solution:

Let $E_{1}$ and $E_{2}$ be the events such that

$E_{1}$ : A speaks truth

$\mathrm{E}_{2}$ : 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 (H) or tail (T).

The probability of getting a head is $\frac{1}{2}$ whether $A$ speaks truth or not

$\therefore \mathrm{P}\left(\mathrm{X} \mid \mathrm{E}_{1}\right)=\mathrm{P}\left(\mathrm{X} \mid \mathrm{E}_{2}\right)=\frac{1}{2}$

The probability that there is actually a head is given by $P\left(E_{1} \mid X\right)$.

$P\left(E_{1} \mid X\right)=\frac{P\left(E_{1}\right) \cdot P\left(X \mid E_{1}\right)}{P\left(E_{1}\right) \cdot P\left(X \mid E_{1}\right)+P\left(E_{2}\right) \cdot P\left(X \mid E_{2}\right)}$

$=\frac{\frac{4}{5} \cdot \frac{1}{2}}{\frac{4}{5} \cdot \frac{1}{2}+\frac{1}{5} \cdot \frac{1}{2}}$

$=\frac{\frac{1}{2} \cdot \frac{4}{5}}{\frac{1}{2}\left(\frac{4}{5}+\frac{1}{5}\right)}$

$=\frac{4}{5}$

$=\frac{4}{5}$

Therefore, the correct answer is A.