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?


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 '15 at 20:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.