I figure this is a trivial question since it's right in the beginning of the book but I get a different answer from that of the answer in the back of the book. I get .0847 while in the correct answer is .0828.

Anyways here is the question:

If birthdays are equally likely to fall on any day, what is the probability that a person chosen at random has a birthday in January?

January has 31 days and there are 365 days in a year so $31 \over 365$ would be $p$ for a non leap year. On a leap year it's $31\over 366$. Since a leap year occurs once every four years I thought I'd get my answer by doing:

$${31\over 365}*{3\over 4} + {31\over 366} * {1\over 4}$$

Any suggestions?


1 Answer 1


Since January has $31$ days, the most days a month can have, and $\frac1{12}= 0.0833\ldots $, there is no obvious way to get a figure as low as $0.0828$.

Either it is a trick question or you have spotted an error.

  • $\begingroup$ ok so it's not just me. thanks $\endgroup$
    – Nolohice
    Jan 27, 2015 at 20:36

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