My Math Forum Can't remember my probability formulas

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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 1-in-250 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 6-sided 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 math-major 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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### how to remember the formular for calculating binomial distribution

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