Find the smallest number of people you need to choose at random so that the probability that at least one has a birthday today exceeds 1/2.

About all I have so far is that (assuming a 365 day calendar year), the probability of a person having a birthday today is 1/365 and the probability of the opposite is 364/365. Where do I go from here?


HINT: Suppose you choose $n$ people. If each person has probability $1-p$ of failing to have a birthday today, what is the probability that all $n$ people fail to have a birthday today?

  • $\begingroup$ by the product rule, wouldn't it be (1-p)^n? (I'm sorry, our professor didn't do a stellar job covering this topic, and our book is a little confusing to read, for me at least) $\endgroup$ – user266529 Oct 26 '15 at 4:42
  • 1
    $\begingroup$ Ah, I see. So it would be N people when 1 - (364/365)^N > 1/2. $\endgroup$ – user266529 Oct 26 '15 at 4:55

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