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` (c) `20/23` (d) `9/20`

Options

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

Answer

C

Explanation

Even G = original signal is green
`E_(1)=A` receives the signal correct
`E_(2)=B` receives the signal correct
E = Signal received by B is green
P (Signal received by B is green)
`=P(GE_(1)E_(2))+P(GbarE_(1)barE_(2))+P(barGE_(1)barE_(2))+P(barGbarE_(1)E_(2))`
`P(E)=(46)/(5xx16)`
`P(G//E)(40//5xx16)/(46//5xx16)=20/23
`