A single which can 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 reciving the signal correctly is `3/4.` If the singal received at station B is green, then the probability that original singal was green is

Options

A. `3/5`
B. `6/7`
C. `20/23`
D. `9/20`

Explanation

To solve the problem, we need to find the probability that the original signal was green given that the signal received at station B is green. We can use Bayes' theorem for this purpose.

### Step-by-Step Solution:

1. **Define Events**:
- Let \( G \) be the event that the original signal is green.
- Let \( R \) be the event that the original signal is red.
- Let \( E \) be the event that the signal received at station B is green.

2. **Given Probabilities**:
- \( P(G) = \frac{4}{5} \)
- \( P(R) = \frac{1}{5} \)
- Probability of receiving the signal correctly at each station = \( \frac{3}{4} \)
- Probability of receiving the signal incorrectly at each station = \( \frac{1}{4} \)

3. **Calculate \( P(E | G) \)**:
- If the original signal is green, the probability that station A receives it correctly and station B receives it correctly is:
\[ P(E | G) = P(E | G, E1, E2) = P(E1) \cdot P(E2) = \frac{3}{4} \cdot \frac{3}{4} = \frac{9}{16} \]

4. **Calculate \( P(E | R) \)**:
- If the original signal is red, the probability that station A receives it incorrectly (transmits green) and station B receives it correctly is:
\[ P(E | R) = P(E1') \cdot P(E2) = \frac{1}{4} \cdot \frac{3}{4} = \frac{3}{16} \]
- The probability that station A receives it correctly (transmits red) and station B receives it incorrectly (transmits green) is:
\[ P(E | R) = P(E1) \cdot P(E2') = \frac{3}{4} \cdot \frac{1}{4} = \frac{3}{16} \]

5. **Total Probability of \( E \)**:
- Using the law of total probability:
\[ P(E) = P(E | G) \cdot P(G) + P(E | R) \cdot P(R) \]
- Substitute the values:
\[ P(E) = \left(\frac{9}{16} \cdot \frac{4}{5}\right) + \left(\frac{3}{16} \cdot \frac{1}{5}\right) = \frac{36}{80} + \frac{3}{80} = \frac{39}{80} \]

6. **Calculate \( P(G | E) \)** using Bayes' theorem:
\[ P(G | E) = \frac{P(E | G) \cdot P(G)}{P(E)} \]
- Substitute the values:
\[ P(G | E) = \frac{\left(\frac{9}{16}\right) \cdot \left(\frac{4}{5}\right)}{\frac{39}{80}} = \frac{\frac{36}{80}}{\frac{39}{80}} = \frac{36}{39} = \frac{12}{13} \]

### Final Answer:
The probability that the original signal was green given that the signal received at station B is green is \( \frac{12}{13} \).