4 ms·

I'm dumb. The probability of rainy Seattle would be (8/27)P(R) where P(R) is the prior probability of rainy weather, no?

1024core 9y ago
Ignoring the P(Rain|Seattle), here's a way to look at this: let's say it's not raining. Then all three must be lying; the probability of that is 1/27. So the probability that it's raining, given that these three are saying it's raining, is 26/27
- chasing 9y ago
  This is my reasoning, as well.
- beebmam 9y ago
  The correct answer. But the question supposes one can know with accuracy what the percent chance that a given statement is a lie, which imo is pretty ridiculous. Fine abstract mathematical problem, but not very useful
- thedirt0115 9y ago
  You CAN'T ignore the extra information like P(Rain|Seattle) though. What if this were the question instead: "Three friends in Seattle told you that Earth exploded and everyone died. Each has a probability of 1/3 of lying. What’s the probability that the Earth exploded and everyone died?"
- bryondowd 9y ago
  Either all three must be lying, or all three must be telling the truth, right? So wouldn't you have to discard the probabilities of them disagreeing? There would be a 1/27 chance of all lying, and wouldn't there only be an 8/27 chance of all telling the truth? The other 18/27 would be a mixture of lies and truths. So really, if we're just comparing the odds of those two outcomes to each other, would the probability just be 8/9 that it is raining in Seattle? Been a while since I've done anything with probabilities, so perhaps I'm way off here.
soVeryTired 9y ago
That doesn't work: if twenty of your friends told you it was raining, your method would tell you the probability was (2/3)^20 P(R), which is very small. Your method was my first thought too - it took a while before I saw how to do it.
OptimalPlay 9y ago
The replies to this don't look correct so here's my take: P(R) = Probability of rain in Seattle P(F) = Probability of three friends saying it's raining in Seattle P(F) = P(F|R)*P(R) + P(F|R')*(1-P(R)) = (8/27)*P(R) + (1/27)*(1-P(R)) = (7/27)*P(R) + (1/27) P(R|F) = (P(F|R)*P(R))/P(F) = [(8/27)*P(R)] / [(7/27)*P(R) + (1/27)] = 8*P(R) / (7*P(R) + 1) So, for instance, if the true probability of rain in Seattle is 50%, the probability of it raining in Seattle given our three friends are saying that it is raining in Seattle is 4/4.5 = 89%

1024core 9y ago

Ignoring the P(Rain|Seattle), here's a way to look at this: let's say it's not raining. Then all three must be lying; the probability of that is 1/27. So the probability that it's raining, given that these three are saying it's raining, is 26/27

chasing 9y ago
This is my reasoning, as well.
beebmam 9y ago
The correct answer. But the question supposes one can know with accuracy what the percent chance that a given statement is a lie, which imo is pretty ridiculous. Fine abstract mathematical problem, but not very useful
thedirt0115 9y ago
You CAN'T ignore the extra information like P(Rain|Seattle) though. What if this were the question instead: "Three friends in Seattle told you that Earth exploded and everyone died. Each has a probability of 1/3 of lying. What’s the probability that the Earth exploded and everyone died?"
bryondowd 9y ago
Either all three must be lying, or all three must be telling the truth, right? So wouldn't you have to discard the probabilities of them disagreeing? There would be a 1/27 chance of all lying, and wouldn't there only be an 8/27 chance of all telling the truth? The other 18/27 would be a mixture of lies and truths. So really, if we're just comparing the odds of those two outcomes to each other, would the probability just be 8/9 that it is raining in Seattle? Been a while since I've done anything with probabilities, so perhaps I'm way off here.

soVeryTired 9y ago

That doesn't work: if twenty of your friends told you it was raining, your method would tell you the probability was (2/3)^20 P(R), which is very small. Your method was my first thought too - it took a while before I saw how to do it.

OptimalPlay 9y ago

The replies to this don't look correct so here's my take: P(R) = Probability of rain in Seattle P(F) = Probability of three friends saying it's raining in Seattle P(F) = P(F|R)*P(R) + P(F|R')*(1-P(R)) = (8/27)*P(R) + (1/27)*(1-P(R)) = (7/27)*P(R) + (1/27) P(R|F) = (P(F|R)*P(R))/P(F) = [(8/27)*P(R)] / [(7/27)*P(R) + (1/27)] = 8*P(R) / (7*P(R) + 1) So, for instance, if the true probability of rain in Seattle is 50%, the probability of it raining in Seattle given our three friends are saying that it is raining in Seattle is 4/4.5 = 89%