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The Facebook question was "What's the chance one of my Facebook friends was born in May?" My answer: Months that are not May = 11. The month that is May = 1, the number of Facebook friends I have = 350. Therefore, P(A) where A=none of my Facebook friends was born in May, P(A)=(11/1)^350. I say that is a probability of 9.15537170E+150/1 So, very likely one of them was born in May.

Am I wrong?

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    $\begingroup$ There seems to be a typographical error: didn't you mean to write "$(11/12)^{350}$" instead of "$(11/1)^{350}$"? The former equals $6\times 10^{-14}$ while the latter is nonsensical. $\endgroup$
    – whuber ♦
    Commented May 8, 2018 at 20:14

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Not quite. Also note that 9.15537170E+150/1=9.15537170E+150 is far, far greater than one. Probabilities by definition are between zero and one (think percentages). i.e. A goalie can't save 500% of the shots against them.

Assuming all of your friends' birthdays are independent of each other (no twins), the probability of any one of them NOT being born in May is approximately $11/12$. That is, if its equally likely to be born in any month, you have a 11 out of 12 chances to be born in not-May.

So, the probability that NONE are born in May is ~$(11/12)^{350}$. That implies that the probability that at least one is born in May is $1-(11/12)^{350}$, which is indeed very likely.

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    $\begingroup$ Also to emphasize the "approximately 11/12" and "if it is equally likely to be born in any month" to the OP, this assumes months are equally likely, which is not the case. Not only do months have different numbers of days, births are not uniformly distributed throughout the year. This approximation is fine for the purposes of this problem, though. $\endgroup$
    – dankernler
    Commented May 8, 2018 at 20:05
  • $\begingroup$ Thank you so much. I don't quite understand your answer. How do I convert 1−(11/12)350 to a percent? Is it 83%??? $\endgroup$
    – John Jackson
    Commented May 8, 2018 at 20:08
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    $\begingroup$ @dankernler Well said! It's actually 99.9999999999941%. $\endgroup$
    – measure_theory
    Commented May 8, 2018 at 20:12
  • $\begingroup$ Much appreciated, now I may be able to sleep tonignt.ha ha $\endgroup$
    – John Jackson
    Commented May 8, 2018 at 20:14

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