Three friends in Tokyo told you that its rainy. Each person has a probability of 1/4 of lying. What is the probability that it is raining in Tokyo? Assume $Pr(rain)$ in Tokyo = 1/5.
Given options:
A) 4/ 101
B) 4/79
C) 2/91
D) 1/109
In my approach I'm getting ans as 27/31.
Pr(y,y,y | rain) * Pr(rain) / (Pr(y,y,y | rain) * Pr(rain) + Pr(y,y,y | no_rain) * Pr(no_rain))
= $(3/4)^3 * 1/5\ /\ ((3/4)^3 * 1/5 + (1/4)^3 * (4/5))$
= $27\ /\ 31$
Where I'm doing mistake?