I've been trying to figure this one out, but I just can't... hopefully someone here can help out There are 25 people in a group, what is the probability that at least one person has their birthday on June 19. Assuming that the probability of being born on a day is 1/365. The answer is .066 Could someone please shed some light on how to do this? I would really appreciate if someone could help me out.

Well, the easiest way to think about this is it's 1-P(zero people with that birthday). So it's 1-(364/365)^25=.066.

Thanks for the help guys. I still don't really get it. But at least I know how to get the answer now. Nick, I'm 17. I'm not actually from Vancouver though, I'm from the BC interior. I just put Vancouver to remain a bit more anonymous on the net I'm in grade 12, and I have my math final this Wednesday!! I get most of it except logs and probability. I'll be going in for help on Tuesday though... Thanks again, Peter

I'm just a 16 year-old with ZERO probabilty experience in math (that I can remember anyway) But why not just 24/365? 24/365 = 0.0657

With all due respect to Chris Derby, 24/365 does round to .066 but it's not the right answer. Here's the logic behind the correct answer. The probability of NOT having June 19 (or any specified day) as your birthday is 364/365. If there were only 2 people in the room, then the probability that neither one was born on 6/19 is 364/365 * 364/365 = (364/365)^2. So the probability of having none of 25 people born on 6/19 is (364/365)^25. The converse of having none born on 6/19 is having at least one born on 6/19 so the answer to your original question is 1-(364/365)^25 = .0662879 The hard way to do it is add up the probability of 1 person with a 6/19 birthday + p(2 people with 6/19 birthday) + ... p(all 25 people with 6/19 birthday).

thats what i meant... i realized that my answer wasn't correct shortly after I posted it and did a little research. its been a while since i've had to do math like that.