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June 5th, 2015, 09:04 PM  #1 
Newbie Joined: Jun 2015 From: New York Posts: 5 Thanks: 0  Can't remember my probability formulas
I'm trying to resolve what I thought was a fairly simple problem, but I haven't managed to get a good answer yet. I want to figure out the odds of getting at least one successful outcome in a series of identical events. I.E., if the odds are 1 in 250, and I make 25 attempts, what are the chances at least one of them will be successful? I'm pretty sure it's not 1/10. I know that if you are trying to figure out the odds of EVERY attempt hitting the 1in250 mark then you multiply the denominators, but that's not what I'm looking for. Someone told me this kind of question is called a cumulative binomial distribution, and I've been looking up info about those via Google but I could still use a hand understanding it in (preferably) simple language. Thanks! 
June 6th, 2015, 03:12 AM  #2 
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361 
Eighter you have 0 successes or at least one, so P(0 successes) + P(at least 1 succes) = 1. This gives P(at least 1 succes) = 1  P(0 successes) P(0 successes) = P(25 failures) = (1  1/250)^25. So P(at least 1 succes) = 1  (1  1/250)^25 ~= 0.09534. Indeed, not 1/10 
June 6th, 2015, 05:50 AM  #3 
Newbie Joined: Jun 2015 From: New York Posts: 5 Thanks: 0 
Thank you very much! You have been extremely helpful. Now if you'd be so kind, let me just make sure I can apply this correctly. Suppose I'm trying to roll a 5 on a 6sided dice. The odds of me getting it in 1 roll are 1 in 6, or ~16.6%. If I make 10 attempts, then my odds of getting at least a single 5 are = 1  (1  1/6)^10 = ~0.8385 = 83.85% Now if I'm attempting to roll either a 4 or a 5, then my chances are 2 in 6 or 1 in 3, so my odds (in 10 attempts) are = 1  (1  1/3)^10 = .9826 = 98.26% If I'm attempting to roll a 3 on 20 sided dice (maybe cause I'm playing D&D or something) in 10 attempts, then its = 1  (1  3/20)^10 = .8032 = 80.32% Please point out where my flaws are, in anywhere. Edit: The formula seems pretty simple now, but do the first two 1's in the equation ever change? would that indicate a different kind of problem? I realize this is sort of a weird way to phrase things, but I'm not a mathmajor and it's been a while since I've taken a class. Last edited by Deepbluediver; June 6th, 2015 at 05:58 AM. 
June 6th, 2015, 08:41 AM  #4 
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361 
Apart from a few rounding errors, I don't see a problem with your work. I think you get it! The ones don't change. The "first" (the one in bold): 1  (1  1/6)^10 is from P(0 successes) + P(at least 1 succes) = 1 The "second" (the one in bold): 1  (1  1/6)^10 is from p(failure) + p(success) = 1. Does that answer your question? Bear in mind that the formula applies for probabilities where is asked for at least 1 succes (and number of successes nonnegative integer). 
June 7th, 2015, 05:37 AM  #5 
Newbie Joined: Jun 2015 From: New York Posts: 5 Thanks: 0 
Thank you very much for your assistance, I think I'm all set at the moment. 

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