How do I solve this?
 

Aside from my final in two weeks, today hopefully marks the last effing math test I'll ever have to take. Here's a homework problem I can't solve:

1. A group of 30 people is selected at random. What is the probability that at least two of them will have the same birthday?

I think this is kind of how it's supposed to be setup:

n(E') = 365P3 = 365!/(365-30)!

That's as far as I can get. Most calculators can't solve such high numbers.

The answer in the back says: 0.71

Re: How do I solve this?
 

Aha! My Discrete and Combinational Math Teacher just did this yesterday!

We had 23 people in our class, and he said it's a half-half probability that two people had the same birthday.

And two people did. It blew my mind.

As for answering the question, don't use a calculator. Use the computer! I'm very sure that programs called Python or Maple would be able to calculate those.

Re: How do I solve this?
 

good luck!

Re: How do I solve this?
 

Pf, that should be very easy.

First I thought of 30/365, then of 29/365, but none of those were 0.71.

Ok, let's compare it with throwing dice. Let's say you throw a dice 2 times. How big is the chance that at least 2 throws are the same? 1/6. Throw it 3 times: (1/6 +1/6) eeeeeeuh crap, it's 0:30 here, I'm tired and I just had a glass of wine. I don't think it matters what the other rolls(/dates) are once you have 1 the same. So no taurenpoopie of double chances or so to speak.

But that would still bring me to 29/365 = 0.079. :unsure:

Blast, I should know this.

Edit: I only looked at the chance of the same as the first. It's also the chances of others being the same. So it's 29/365 + 28/365 + 27/365..... = ???

Edit3: I probably made another mistake, because that gives a probability of 1.09 something.

Re: How do I solve this?
 

Maple.. :errr:

Re: How do I solve this?
 

Ignoring the possibility that someone was born on Feb29...


P = (365)(364)(363)....(365 - n + 1)/(365)^n where n = 30.

P = .29368, which is actually the probability that no two people have the same birthday, so

1 - .29368 = .70632 => .71

So in a room with 30 people, chances are more likely than not that two people will have the same birthday. May seem counterintuitive but it's true.

Re: How do I solve this?
 

Don't feel bad, it took mathematicians a long time to figure out how to calculate those kinds of probabilities.

Re: How do I solve this?
 

I hate probabilities :p I did one semester of Statistics and it neary did my head in, heh.

Re: How do I solve this?
 

Thanks for the help and encouraging words, my friends. Luckily, there was no problem this challenging on the test. My teacher is cool, he'll teach us the concepts, and then the homework in the math book features normal to difficult problems. And then on the test, he always makes them as simple and straight-forward as possible.

Re: How do I solve this?
 

I wish my Math teachers were like that.

Sigh.