Hello, I am bugfree Assistant. Feel free to ask me for any question related to this problem
You have three friends in Seattle, each with a 2/3 chance of telling the truth and a 1/3 chance of lying. All three friends tell you it is raining. You need to determine the probability that it is actually raining.
Understanding the Problem:
Calculate the Probability of All Friends Lying:
If all friends are lying, then it is not raining.
Probability that a single friend lies: 1/3.
Probability that all three friends lie:
P(All Lie)=(31)3=271
Calculate the Probability of Raining:
The probability it is not raining is the probability that all friends are lying.
Thus, the probability it is raining is:
P(Rain)=1−P(All Lie)=1−271=2726
Assumptions:
Calculate Conditional Probabilities:
Probability all friends say it is raining given it is raining (all tell the truth):
P(F1∩F2∩F3∣R)=(32)3=278
Probability all friends say it is raining given it is not raining (all lie):
P(F1∩F2∩F3∣Rc)=(31)3=271
Apply Bayes' Theorem:
We use Bayes' theorem to find the probability it is raining given all friends say it is raining:
P(R∣F1∩F2∩F3)=P(F1∩F2∩F3)P(F1∩F2∩F3∣R)⋅P(R)
Calculate Total Probability:
Total probability that all friends say it is raining:
P(F1∩F2∩F3)=P(F1∩F2∩F3∣R)⋅P(R)+P(F1∩F2∩F3∣Rc)⋅P(Rc)
=278⋅0.5+271⋅0.5=548+541=549=61
Final Calculation:
Substitute back into Bayes' theorem:
P(R∣F1∩F2∩F3)=61278⋅0.5=98
The Bayesian approach yields a probability of 98 that it is raining given all friends say it is raining, under the assumption of a 50% prior probability of rain. The frequentist approach, focusing on the probability that at least one friend is truthful, gives a probability of 2726. The difference arises due to the assumptions and focus of each method.