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.
I count only three transmissions: A to B; B to C and C to D; hence raised the transmission matrix to the third power; the probabilities at A were the original probabilities:
The transition matrix, in the order green, yellow and red, for one transmission is:
[5/7 1/7 1/7;
1/7 5/7 1/7;
1/7 1/7 5/7]
Then, for three transmissions the matrix is the original one cubed:
[157/343, 93/343, 93/343]
[ 93/343, 157/343, 93/343]
[ 93/343, 93/343, 157/343]
Considering only the 5/7 that were originally green, after three transmissions that would come out
[4/7 0 0] * [157/343, 93/343, 93/343]
[ 93/343, 157/343, 93/343]
[ 93/343, 93/343, 157/343]
= [628/2401, 372/2401, 372/2401]
Overall, the final probabilities including the contributions of yellow and red as the initial observation:
[4/7 2/7 1/7] * [157/343, 93/343, 93/343]
[ 93/343, 157/343, 93/343]
[ 93/343, 93/343, 157/343]
= [907/2401, 779/2401, 715/2401]
So the probability the original signal was green was 628/907 =~ 0.69239250275634.
Matrix operations performed by this program:
transMatrix=sym([5/7 1/7 1/7;
1/7 5/7 1/7;
1/7 1/7 5/7])
tm3=transMatrix^3
[1 0 0]*tm3
[4/7 0 0]*tm3
[4/7 2/7 1/7]*tm3
Edited on April 5, 2023, 5:01 pm
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Posted by Charlie
on 2023-04-05 16:59:39 |
solution
