Bayes Theorem Probability of Signal Transmission

To find the probability that the original signal was green given that the signal received at station B is green, we apply Bayes' Theorem.

Step 1: Define the Events

Let:

• $E_1$: Event that the original signal transmitted is Green.
• $E_2$: Event that the original signal transmitted is Red.
• $B_G$: Event that the signal received at station B is Green.

From the question, the initial probabilities are:
$P(E_1) = \frac{4}{5}$
$P(E_2) = \frac{1}{5}$

The probability of any station receiving and transmitting the signal correctly is $P(\text{Correct}) = \frac{3}{4}$, and the probability of error is $P(\text{Error}) = 1 - \frac{3}{4} = \frac{1}{4}$.

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Step 2: Compute Conditional Probabilities

Case 1: Probability that station B receives Green given that the original signal was Green, $P(B_G \mid E_1)$

The signal received at B will be Green under two conditions:

1. Both station A and station B receive the signal correctly:
   $P(\text{Correct at A}) \times P(\text{Correct at B}) = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16}$

2. Both station A and station B make an error:
   Station A receives Green as Red (error), and then Station B receives Red as Green (error).
   $P(\text{Error at A}) \times P(\text{Error at B}) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$

Adding these mutually exclusive possibilities:
$P(B_G \mid E_1) = \frac{9}{16} + \frac{1}{16} = \frac{10}{16}$

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Case 2: Probability that station B receives Green given that the original signal was Red, $P(B_G \mid E_2)$

The signal received at B will be Green under two conditions:

1. Station A makes an error (Red $\rightarrow$ Green) and Station B receives correctly (Green $\rightarrow$ Green):
   $P(\text{Error at A}) \times P(\text{Correct at B}) = \frac{1}{4} \times \frac{3}{4} = \frac{3}{16}$

2. Station A receives correctly (Red $\rightarrow$ Red) and Station B makes an error (Red $\rightarrow$ Green):
   $P(\text{Correct at A}) \times P(\text{Error at B}) = \frac{3}{4} \times \frac{1}{4} = \frac{3}{16}$

Adding these mutually exclusive possibilities:
$P(B_G \mid E_2) = \frac{3}{16} + \frac{3}{16} = \frac{6}{16}$

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Step 3: Apply Bayes' Theorem

We want to find $P(E_1 \mid B_G)$, the conditional probability that the original signal was Green given that the final signal received at B is Green:

$P(E_1 \mid B_G) = \frac{P(B_G \mid E_1) \cdot P(E_1)}{P(B_G \mid E_1) \cdot P(E_1) + P(B_G \mid E_2) \cdot P(E_2)}$

Substitute the calculated values into the formula:

$P(E_1 \mid B_G) = \frac{\left(\frac{10}{16}\right) \times \left(\frac{4}{5}\right)}{\left(\frac{10}{16}\right) \times \left(\frac{4}{5}\right) + \left(\frac{6}{16}\right) \times \left(\frac{1}{5}\right)}$

$P(E_1 \mid B_G) = \frac{\frac{40}{80}}{\frac{40}{80} + \frac{6}{80}} = \frac{40}{46} = \frac{20}{23}$

Thus, the required probability is $\frac{20}{23}$.

Correct Answer: C ($\frac{20}{23}$)