A single which can can be green or red with probability `2/3 and 1/5` respectively, is received by station A and then transmitted to station B. The probability of each station reciving the signal correctly is `3/4.` If the singal received at station B is green, then the probability that original singal was green is

Options

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

Answer

C

Explanation

It is evident from the tree diagram that
`P(B_(G))=(4)/(5)xx(3)/(4)xx(3)/(4)+(4)/(5)xx(1)/(4)xx(1)/(4)+(1)/(5)xx(3)/(4)xx(1)/(4)+(1)/(5)xx(1)/(4)xx(1)/(4)xx(3)/(4)=(23)/(40)`
`P((B_(G))/(G))=(3)/(4)xx(3)/(4)+(1)/(4)xx(1)/(4)=(5)/(8)`
`implies P(B_(G) cap G)=(4)/(5)xx(5)/(8)=(1)/(2)`
`therefore` Required probability `=P((G)/(B_(G)))=(P(B_(G) cap G))/(P(B_(G)))`
`=((1)/(2))/((23)/(40))=(20)/(23)`