A signal can be green, yellow, or red with respective probabilities 4/7, 2/7, and 1/7, is received by station A and transmitted to station B, and station B then transmits the signal to station C and, finally station D receives the signal from station C.
The probability of each station receiving the signal correctly is 5/7,and that if it's received incorrectly it is equally likely to be received as either of the two other colors.
Determine the probability that the original signal was green, given that the signal received by station D is green.
So I started by making a transition matrix that takes in a vector for probabilities of r/y/g of a station changing a signal it receives.
[ 5/7 1/7 1/7 ]
[ 1/7 5/7 1/7 ]
[ 1/7 1/7 5/7 ]
Raising this to the fourth power represents the cumulative effect of all four transmitters.
[ 971/2401 715/2401 715/2401 ]
[ 715/2401 971/2401 715/2401 ]
[ 715/2401 715/2401 971/2401 ]
But then I noticed we can just read the probabilities right off this matrix: The probability a signal came through D the same as the original prior to A is 971/2401 and the probability a signal came through changed is 715/2401.
Then from the given values we have (4/7)*(971/2401)=3884/16807 chance of a signal starting green and ending green.
And we have a (2/7)*(715/2401)=1430/16807 chance of a signal starting yellow and ending green.
And we have a (1/7)*(715/2401)=715/16807 chance of a signal starting red and ending green.
So then the probability that the original signal was green, given that the signal received by station D is green equals
(3884/16807) / (3884/16807+1430/16807+715/16807) = 3884/6029 ~= 64.42%.
Solution
