Question : A signal which 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 receiving the signal correctly is 3/4. If the signal received at station B is green, then the probability that the original signal green is
(A) 3/5
(B) 6/7
(C) 20/23
(D) 9/20
Answer : option (C)
Solution :
From the three diagram it follows that
=P\left( G| Bg\right) =\dfrac{1}{2}\times \dfrac{80}{46}=\dfrac{40}{46}=\dfrac{20}{23}