(adsbygoogle = window.adsbygoogle || []).push({}); Weird probability problem..."With what probability is it raining?"

Came across this the other day looking at interview questions.

Suppose you want to determine whether it's raining in a distant city. You have three friends there who you can call and ask about this. Only thing is:

So, if you call all your friends, and they all tell you it's raining...what's the probability it's actually raining?

- Each friend will tell the truth with probability 2/3.
- Each friend will lie with probability 1/3.
- The event that Friend [itex]i[/itex] lies is independent of the event that Friend [itex]j[/itex] lies for [itex]1 \leq i,j \leq k[/itex].

Here's a naive answer: The probability in question is just the probability that at least one of them is telling the truth, which by independence is [itex]P(t_1 \cup t_2 \cup t_3) = 1 - P(\ell_1 \cap \ell_2 \cap \ell_3) = 1 - P(\ell_1)P(\ell_2)P(\ell_3) = 1 - (1/3)^3 = 26/27[/itex]. But there is a conceivable objection to this: You don't needat least oneof them to be telling the truth; you need themALLto be telling the truth. Because the probability just computed includes, for instance, the event that Friend 1 is telling the truth (it's raining), but Friend 2 and 3 are lying (it's not raining), which isincoherent. So, in a sense, the sample space used in the calculation above is too big! And what you should compute instead is [itex]P(t_1 \cap t_2 \cap t_3) = P(t_1)P(t_2)P(t_3) = (2/3)^3 = 8/27[/itex]. But...that doesn't quite make sense...where's the remaining [itex]1 - 1/27 - 8/27 = 2/3[/itex] of our probability measure living?

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# Weird probability problem With what probability is it raining?

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