Question

A signal which can be green or red with probability $\frac{4}{5}$ and $\frac{1}{5}$ respectively is recieved 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 recieved at station B is green, then the probability that the original signal was green is

• A. $3$
• B. $\frac{6}{7}$
• C. $\frac{20}{23}$
• D. $\frac{9}{20}$

Answer 1

The correct option is C $\frac{20}{23}$

P(signal at B is green) (IV cases possible)
P(original signal is green) $\times$ P(it is received correctly at A and B ) +
P(original signal is red) $\times$ P(it is received wrongly at A but correctly at B) +
P(original signal is red) $\times$ P(it is received wrongly at B but correctly at A)
+P(original signal is green) $\times$ P(it is received wrongly at both A and B )
$=\frac{4}{5}\times (\frac{3}{4}{)}^2+\frac{1}{5}\times \frac{3}{4}.\frac{1}{4}.+\frac{1}{5}\times \frac{3}{4}.\frac{1}{4}+\frac{4}{5}\times (\frac{1}{4}{)}^2$
$=\frac{46}{80}$
P(original signal is green given received at B is green) = $\frac{40/80}{46/80}=\frac{20}{23}$
Sum of Case 1 + Case IV