A signal which can be green or red with probability `4/5 and 1/5` respectively, is received by station A and then and 3 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 was green is (a) `3/5` (b) `6/7` (d) `20/23` (d) `9/20`

Explanation

acc to the question

`P = (4/5*3/4*3/4 + 4/5*1/4*1/4) + (1/5*3/4*1/4 + 1/5*1/4*3/4)`

`P = 46/80 = 23/40`

`P [or G] = 40/80 = 20/40`

`P= (20/40)/(23/40)= 20/23`