A person is known to speak truth 3 out of 4 times. He  throws a dice and responds that it is a six. Then the  probability that it is actually a six is

Let
E = the event that A reports it is six

 E₁ = six turns up when a dice is thrown 

E₂ = six does not turns up when a dice is thrown Probability that six actually showed up if A reported that of was six $=P\left(E_1/E\right)$

$\begin{array}{l}P\left(E_1/E\right)=\frac{P\left(E_1\cap E\right)}{P\left(E\right)}\\ =\frac{P\left(E_1\right)\cdot P\left(E/E_1\right)}{P\left(E_1\right)\cdot P\left(E/E_1\right)+P\left(E_2\right)\cdot P\left(E/E_2\right)}\end{array}$

$P\left(E/E_1\right)=$Probability that the person reports six when it actually showed 

$P\left(E/E_2\right)=$ Probability that the person reports six when it did not show six 

$=\text{ Probability that he lies }=1-\frac{3}{4}$

$=\frac{1}{4}$

$\text{ also }P\left(E_1\right)=\frac{1}{6}\text{ and }P\left(E_2\right)=\frac{5}{6}$

$P\left(E_1/E\right)=\frac{\frac{1}{6}\cdot \frac{3}{4}}{\frac{1}{6}\cdot \frac{3}{4}+\frac{5}{6}\cdot \frac{1}{4}}=\frac{3}{8}$