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

$P(G) = \dfrac{4}{5}, \qquad P(R) = \dfrac{1}{5}$

$P(\text{correct at each station}) = \dfrac{3}{4}$

$\text{To find: } P(\text{original green} \mid \text{B receives green})$

$\text{each station flips colour with prob } 1 - \dfrac{3}{4} = \dfrac{1}{4}$

$\text{so from original to B: } 0,\ 1 \text{ or } 2 \text{ flips}$

$1.\ P(\text{B green} \mid G),\ P(\text{B green} \mid R)$

$2.\ P(\text{B green}) \text{ (total probability)}$

$3.\ \text{Bayes: } P(G \mid \text{B green}) = \dfrac{P(G)\,P(\text{B green}\mid G)}{P(\text{B green})}$

$G \to \text{B green: } 0 \text{ or } 2 \text{ flips (even)}$

$G \to \text{B red: } 1 \text{ flip (odd)}$

$P(\text{BG} \mid G) = \left(\dfrac{3}{4}\right)^2 + \left(\dfrac{1}{4}\right)^2 = \dfrac{9}{16} + \dfrac{1}{16} = \dfrac{10}{16} = \dfrac{5}{8}$

$P(\text{BG} \mid R) = 2 \cdot \dfrac{3}{4}\cdot\dfrac{1}{4} = \dfrac{6}{16} = \dfrac{3}{8}$

$P(\text{BG}) = P(G)P(\text{BG}\mid G) + P(R)P(\text{BG}\mid R)$

$= \dfrac{4}{5}\cdot\dfrac{5}{8} + \dfrac{1}{5}\cdot\dfrac{3}{8} = \dfrac{1}{2} + \dfrac{3}{40} = \dfrac{23}{40}$

$P(G \mid \text{BG}) = \dfrac{P(G \cap \text{BG})}{P(\text{BG})} = \dfrac{P(G)\,P(\text{BG}\mid G)}{P(\text{BG})}$

$= \dfrac{\dfrac{4}{5}\cdot\dfrac{5}{8}}{\dfrac{23}{40}} = \dfrac{\dfrac{1}{2}}{\dfrac{23}{40}}$

$= \dfrac{1}{2} \times \dfrac{40}{23} = \dfrac{20}{23}$

$P(G \mid \text{BG}) = \dfrac{20}{23}$

$\boxed{\text{Answer: (3)}\ \ \dfrac{20}{23}}$