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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 pr
by 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 agoThis is my reasoning, as well.
- beebmam 9y agoThe 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 agoYou 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 agoEither 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.