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

Options

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

Answer

C

Explanation

From the tree-diagram it follows that,
`P(B_(G))=(4)/(5)((3)/(4)xx(3)/(4)+(1)/(4)xx(1)/(4))+(1)/(5)((3)/(4)xx(1)/(4)+(1)/(4)xx(3)/(4))=(46)/(80)`

`P(B_(G)//G)=((3)/(4)xx(3)/(4)+(1)/(4)xx(1)/(4))=(10)/(16)=(5)/(8)`
`therefore P(B_(G)nn G)=(5)/(8)xx(4)/(5)=(1)/(2)`
`P(G//B_(G))=(P(B_(G)nn G))/(P(B_(G)))=(1)/(2)xx(80)/(46)=(20)/(23)`