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 think this is correct. I don't have time to double check the logic.
If I read the problem correctly, there are four received transmissions.
Let's start by finding the probability a signal gets where it's going.
The mapping (g,y,r) -> (5g+y+r, g+5y+r, g+y+5r)/7
is the reception in terms of the amounts sent.
A green signal sent four times becomes
(1,0,0)/7
-> (5,1,1)/7^2
-> (27,11,11)/7^3
-> (157,93,93)/7^4
-> (971,715,715)/7^5
the other colors are just a permutation of these.
The chance that a signal starts and ends green is then
4/7 * 971/7^5 = 3884/7^6
The chance that a signal starts yellow and ends green is
2/7 * 715/7^5 = 1439/7^6
The chance that a signal starts red and ends green is
1/7 * 715/7^5 = 715/7^6
Final answer 3884/(3884+1439+715) = 3884/6038 = 1942/3019
or about 64.3%
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Posted by Jer
on 2023-04-05 13:41:32 |
Solution
