A slip of paper is given to a person `A` who marks it either with a plus sign or a minus sign. The probability of his writing a plus sign is `1//3`. A passes the slip to `B`, who may either leave it alone or change the sign before passing it to `C`. Next `C` passes the slip to `D` after perhaps changing the sign. Finally `D` passes it to a refere after perhaps changing the sign. `B,C,D each change the sign with probability `2//3`.
If the refree observes a plus sign on the slip then the probability that A originally wrote a plus sign is

Options

A. `13//41`
B. `19//27`
C. `17//25`
D. `21//37`

Answer

A

Explanation

`(a)` Let `E_(1)=` Event that `A` wrote a plus sign
`E_(1)=` Event that `B` wrote a plus sign
`E=` Event that the referee observes a plus sign.
Given `P(E_(1))=(1)/(3)impliesP(E_(2))=(2)/(3)`
`P(E//E_(1))=` Probability that none of `B`, `C`, `D` change sign `+` Probability that exactly two of `B,C,D` change sign
`=(1)/(27)+3((1)/(3)xx(2)/(3)xx(2)/(3))=(13)/(27)`
`P(E//E_(2))=` Probability that all of `B,c,d` change sign `+` Probability that exactly one of `B,C,D` changes sign ,brgt `=(8)/(27)+3((2)/(3)xx(1)/(3)xx(1)/(3))=(14)/(27)`
`:.P(E_(1)//E)=(13)/(41)` (Using Bayes' Theorem)