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Chance of something happening. 
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#1
May2608, 07:45 PM

P: 863

This has bugged me for quite some time now, but I've always shrugged it off. I've taken a course in statistics and have dealt with them in one way or another for a few years, so I KNOW I've seen the answer to this, but I can't remember it and don't know where to look for it.
Well, I could crack open my statistics book, but I figure this would go faster. Anyway, my question is as follows: The probability of something happening is 1/100. You do 100 iterations. What is the chance of it happening? So for example you have a 100 sided die, and you roll it 100 times. What's the chance of getting a 9? I know I can't be 1. But it has to get closer as you get to infinity, right? I just can't think off the top of my head how that formulation would go. 


#2
May2608, 08:28 PM

P: 279

What is the chance of getting exactly one nine or at least one nine?
I think at least one nine is 1.99^100 


#3
May2608, 08:54 PM

P: 863

That makes sense. And yeah, I meant "at least", but "exactly" would be a good one to know, too. Thanks.



#4
May2608, 10:43 PM

P: 279

Chance of something happening.
I had to double check this one:
Exactly one nine is C(100,1)(.01^1)(.99^99) then exactly two nines C(100,2)(.01^2)(.99^98) and so on... 


#5
May2708, 12:25 AM

P: 863

Oh, duh, it's a combination.
Okay, but do you have a quick derivation for getting at least 1 number? It's not obvious to me how you would get that from seeing the formula, but I suspect it has something to do with combinations anyway. 


#6
May2708, 01:51 AM

P: 279

I didn't derive it. I just considered not no nines. So the probability of any result of 100 rolls (1) minus the probability of all the rolls being any number but nine (99^100). In this case nine, but it doesn't matter. The probability is the same for at least one of any single number 1100.
I suppose you could sum the series of combinations. But that would be work I think. 


#7
May2708, 04:05 PM

P: 863

Ahh I gotcha. Thanks. :)



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