A signal which can be green or red with probability $$\frac{4}{5}$$ and $$\frac{1}{5}$$ respectively, is received by station $$A$$ and then transmitted to station $$B.$$ The probability of each station receiving the signal correctly is $$\frac{3}{4} .$$ If the signal received at station $$B$$ is green, then the probability that the original signal was green is

Solution :
Let $$G \equiv $$  original signal is green
⇒ $$P(G) = \frac{4}{5}$$
$${E_1}$$ $$ \equiv $$ $$A$$ receives the signal correctly $$P\left( {{E_1}} \right) = \frac{3}{4}$$
$${E_2}$$ $$ \equiv $$ $$B$$ receives the signal correctly $$P\left( {{E_2}} \right) = \frac{3}{4}$$
$$E \equiv $$  Signal received by $$B$$ is green.
Then $$E$$ can happen in the following ways

Original Signal	Received at A	Received at B
Red $$ \to $$	Red $$ \to $$	Green
Red $$ \to $$	Green $$ \to $$	Green
Green $$ \to $$	Green $$ \to $$	Green
Green $$ \to $$	Red $$ \to $$	Green

$$\eqalign{ & \therefore \,\,P\left( E \right) = \left( {\overline G \cap {E_1} \cap \overline {{E_2}} } \right) + \left( {\overline G \cap {E_1} \cap \overline {{E_2}} } \right) + P\left( {G \cap {E_1} \cap {E_2}} \right) + P\left( {G \cap \overline {{E_1}} \cap \overline {{E_2}} } \right) \cr & = \frac{1}{5} \times \frac{3}{4} \times \frac{1}{4} + \frac{1}{5} \times \frac{3}{4} \times \frac{1}{4} + \frac{4}{5} \times \frac{3}{4} \times \frac{3}{4} + \frac{4}{5} \times \frac{1}{4} \times \frac{1}{4} \cr & = \frac{{3 + 3 + 36 + 4}}{{80}} \cr & = \frac{{46}}{{80}} \cr & = \frac{{23}}{{40}} \cr} $$
$$\eqalign{ & \therefore \,\,P\left( {\frac{G}{E}} \right) = \frac{{P\left( {G \cap E} \right)}}{{P\left( E \right)}} \cr & = \frac{{P\left( {G \cap {E_1} \cap {E_2}} \right) + P\left( {G \cap {{\overline E }_1} \cap {{\overline E }_2}} \right)}}{{P\left( E \right)}} \cr & = \frac{{\frac{4}{5} \times \frac{3}{4} \times \frac{3}{4} + \frac{4}{5} \times \frac{1}{4} \times \frac{1}{4}}}{{\frac{{23}}{{40}}}} \cr & = \frac{{\frac{{40}}{{80}}}}{{\frac{{23}}{{40}}}} \cr & = \frac{{20}}{{23}} \cr} $$