<p>ok, someone please help me with this math problem. If there are 40 kids in my class, what is the probability that 2 have the same birthday? (please show me how you did this)</p>
<p>Compute the probability that all of them have different birthdays. Since this is equivalent to NOT(at least two of them have the same birthday), taking 1 - the number you get gives the answer.</p>
<p>The probability that all of them have different birthdays is (for obvious reasons):</p>
<p>365/365<em>364/365</em>363/634<em>. . .</em>(365-40+1)/365</p>
<p>you get to pick any of 365 days for the first one, 364 days for the next one (any but the one already taken), etc.</p>
<p>Taking 1 minus that product is the answer.</p>
<p>And don't post math questions on CC! ;-)</p>
<p>[I suppose I am giving the incorrect incentives by answering them. So much for all that economics training.]</p>
<p>but well the problem could have been ,..</p>
<p>What is the probability that exactly 2 have the same birthday? </p>
<p>This one definitely would be a more trickier problem to solve..</p>
<p>SM</p>
<p>If it's exactly 2, then Ben's solution is slightly incorrect (sorry, don't hurt me).</p>
<p>take the number of possibilities that produce 2 kids with the same birthday = (40 C 2)<em>365 and multiply it by the possibilities for the other kids, i.e. 364</em>363<em>362....</em>327 = (364!)/(326!). </p>
<p>Then, take all of that and divide by total number of possibilities, 365^40. </p>
<p>BTW, Art of Problem Solving is much better for answering math questions than CC (not to rag on Ben's math skills).</p>
<p>:) I knew that I gave the solution to the problem the OP wanted to ask about, not the one he actually asked about. If the question is "exactly 2", asiaknight is right, but I'd bet money that the question on the homework/whatever is the classic "2 or more".</p>
<p>And yes, asiaknight, I've been referring people to AOPS for a while now : )</p>