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snarescience.com • Probability Question: Really *beep* Hard
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### Probability Question: Really *beep* Hard

Posted: Wed Feb 29, 2012 10:05 pm
So, there are 100 freshmen in my dorm. In room pick season, freshmen are given a lottery number 1-100.

A group of 4 freshmen wish to get a quad. Only one of them needs to get a high number to get the room they want. Let's cut it at 25.

So, what's the probability that any one of the four students gets a number in the top 25? I've been struggling with this for a few minutes now.

I figure its got to be better than 25%, because each persons chances get better and better, but I don't know how to calculate that.

Make me feel dumb.

### Re: Probability Question: Really *beep* Hard

Posted: Wed Feb 29, 2012 11:10 pm
Exactly one of them get a top 25 number or at least one of them get a top 25?

### Re: Probability Question: Really *beep* Hard

Posted: Wed Feb 29, 2012 11:15 pm
P(at least one in the top 25) = 1 - P(none in top 25)

1-P(none in top 25) = 1-(75/100)*(74/99)*(73/98)*(72/97)

I think that's right. Paging Kevin Troyanos!

### Re: Probability Question: Really *beep* Hard

Posted: Thu Mar 01, 2012 12:11 am
I was trying to figure out how to figure out at least one getting in the top 25.

And that's the same number I calculated with the help of someone else. Its about 31%, which sounds reasonable.

### Re: Probability Question: Really *beep* Hard

Posted: Thu Mar 01, 2012 10:21 am
*beep* statistics

### Re: Probability Question: Really *beep* Hard

Posted: Thu Mar 01, 2012 5:07 pm
snarescience wrote:P(at least one in the top 25) = 1 - P(none in top 25)

1-P(none in top 25) = 1-(75/100)*(74/99)*(73/98)*(72/97)

I think that's right. Paging Kevin Troyanos!
I think that equation would assume that those 4 would be the first four picked, and not necessarily at random.
I could be wrong though.
It's been forever since I've done probablity/math in general.

### Re: Probability Question: Really *beep* Hard

Posted: Thu Mar 01, 2012 5:13 pm
Andymac wrote:
snarescience wrote:P(at least one in the top 25) = 1 - P(none in top 25)

1-P(none in top 25) = 1-(75/100)*(74/99)*(73/98)*(72/97)

I think that's right. Paging Kevin Troyanos!
I think that equation would assume that those 4 would be the first four picked, and not necessarily at random.
I could be wrong though.
It's been forever since I've done probablity/math in general.
25X25X25X25 = 390,625%

### Re: Probability Question: Really *beep* Hard

Posted: Thu Mar 01, 2012 6:16 pm
Andymac wrote:
snarescience wrote:P(at least one in the top 25) = 1 - P(none in top 25)

1-P(none in top 25) = 1-(75/100)*(74/99)*(73/98)*(72/97)

I think that's right. Paging Kevin Troyanos!
I think that equation would assume that those 4 would be the first four picked, and not necessarily at random.
I could be wrong though.
It's been forever since I've done probablity/math in general.
If everyone picked blindly, this math should work out correctly regardless of the order picked. If you weren't first to pick and had knowledge of which numbers had already been picked, then yes, the math would be different.

### Re: Probability Question: Really *beep* Hard

Posted: Thu Mar 29, 2012 7:40 pm
each person has a 1/4 chance of getting in the top 25.

### Re: Probability Question: Really *beep* Hard

Posted: Fri Mar 30, 2012 6:09 pm
snarescience wrote:
Andymac wrote:
snarescience wrote:P(at least one in the top 25) = 1 - P(none in top 25)

1-P(none in top 25) = 1-(75/100)*(74/99)*(73/98)*(72/97)

I think that's right. Paging Kevin Troyanos!
I think that equation would assume that those 4 would be the first four picked, and not necessarily at random.
I could be wrong though.
It's been forever since I've done probablity/math in general.
If everyone picked blindly, this math should work out correctly regardless of the order picked. If you weren't first to pick and had knowledge of which numbers had already been picked, then yes, the math would be different.
This