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I came across following problem in probability:

Simon wanted to mail one of his friends. He had two email IDs, but only one of them was correct. He was not sure which one. So to check which one, he immediately mailed one of the IDs. But he did not get any mail undelivered warning, so he concluded that it was correct email ID. Assuming that email server will fail to send undelivered warning for an email targeting to non existing email IDs at any given time with a probability of 1%, what is the probability that Simon's conclusion was wrong?

I got confused with what can be exact approach. I have following different solutions:

  • Since Simon already did not get any mail undelivered warning, we do not need to consider 1% probability and that now it all boils down to whether even after no undelivered (non existence email) warning, is it correct email id or incorrect one. So there are only two possibilities. Thus the probability that Simon's conclusion is wrong is 0.5

  • However, then I reread the problem and "at any given time with a probability of 1%" confused me further and I thought I should really be multiplying 0.5 by 1% to get 0.05 or 5%.

  • But then I had another thought that we cannot multiply this straight, as when we get undelivered message, there is no possibility that the mail belonged to Simon's friend. So for those 99%, the mail will surely not belong to Simon's friend. So we have 101 possibilities:

    • 99 possibilities (for which we get undelivered warning) will surely conclude that the mail do not belong to Simon' and
    • 1 possibility in which we get undelivered warning and mail was really non existent
    • 1 possibility in which we get undelivered warning and mail was indeed for Simon's friend

    So desired probability will be 1 divided by 101 which is approx 0.0099.

Which one is correct? I intuitively feel the last one is more apt, but not really having clear / mathematical logic why so (if that was indeed the correct solution).

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  • $\begingroup$ "Almost" your second solution. Formally, a routine application of Bayes' Thm. You seek $P(C|W^c)=P(CW^c)/P(W^c).$ Num: $P(C)P(W^c|C) = (1/2)(1)= 0.5.$ Denom: $P(C)P(W^c|C) + P(C^c)P(W^c|C^c) = (1/2)+(1/2)(.01) = 0.505.$ Ans: $0.5/0.505 =100/101= 0.990099.$ $\endgroup$ – BruceET Aug 23 '20 at 20:58

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