1. Define events

Let:

• $G$ = original signal is green
• $R$ = original signal is red
• $B_G$ = station $B$ receives green

Given: $P(G)=\frac45, \qquad P(R)=\frac15$

Each station receives/transmits correctly with probability: $\frac34$ So, each station makes an error with probability: $\frac14$

We need to find: $P(G\mid B_G)$ using Bayes' theorem.

2. Find $P(B_G\mid G)$

If the original signal is green, then for station $B$ to finally receive green, two possible paths exist:

• $A$ receives green correctly and transmits it correctly to $B$: $\frac34\cdot \frac34 = \frac{9}{16}$

• $A$ receives it wrongly as red, and then $B$ receives that wrongly as green: $\frac14\cdot \frac14 = \frac{1}{16}$

Hence, $P(B_G\mid G)=\frac{9}{16}+\frac{1}{16}=\frac{10}{16}=\frac58$

3. Find $P(B_G\mid R)$

If the original signal is red, then for station $B$ to receive green, two possible paths exist:

• $A$ receives red correctly, but $B$ receives it wrongly as green: $\frac34\cdot \frac14 = \frac{3}{16}$

• $A$ receives red wrongly as green, and $B$ receives green correctly: $\frac14\cdot \frac34 = \frac{3}{16}$

Thus, $P(B_G\mid R)=\frac{3}{16}+\frac{3}{16}=\frac{6}{16}=\frac38$

4. Find total probability that $B$ receives green

By total probability, $P(B_G)=P(B_G\mid G)P(G)+P(B_G\mid R)P(R)$

So, $P(B_G)=\frac58\cdot \frac45 + \frac38\cdot \frac15$

$P(B_G)=\frac12 + \frac{3}{40}=\frac{20}{40}+\frac{3}{40}=\frac{23}{40}$

5. Apply Bayes' theorem

$P(G\mid B_G)=\frac{P(B_G\mid G)P(G)}{P(B_G)}$

Substitute values: $P(G\mid B_G)=\frac{\frac58\cdot \frac45}{\frac{23}{40}}$

$P(G\mid B_G)=\frac{\frac12}{\frac{23}{40}}=\frac12\cdot \frac{40}{23}=\frac{20}{23}$

6. Match with options

$\frac{20}{23}$ corresponds to Option C.

7. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So, the answer agrees with the stored correct answer.