**Question **: A signal which can be green or red with probability 4/5 and 1/5 respectively, is received by station A and then transmitted to station B. The probability of each station receiving the signal correctly is 3/4. If the signal received at station B is green, then the probability that the original signal green is

(A) 3/5

(B) 6/7

(C) 20/23

(D) 9/20

**Answer **: option (C)

**Solution **:

From the three diagram it follows that

\begin{aligned}\Rightarrow P\left( Bg\right) =\dfrac{46}{80}\ \Rightarrow P\left( B_{g}| G\right) =\dfrac{10}{16}=\dfrac{5}{8}\ \Rightarrow P\left( B_{g}\cap G\right) =\dfrac{5}{8}\times \dfrac{4}{5}=\dfrac{1}{2}\end{aligned}=P\left( G| Bg\right) =\dfrac{1}{2}\times \dfrac{80}{46}=\dfrac{40}{46}=\dfrac{20}{23}