07302009, 08:03 AM



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if probability of failure is X
what is the probability of N failures in Q attempts?

07302009, 08:35 AM



Location: Londonderry, NH
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How about asking how long until the first failure?

07302009, 08:40 AM



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Quote:
Originally Posted by GregW
How about asking how long until the first failure?

???
i need to know what the probability of N failures in Q attempts?

07302009, 10:49 AM



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Don't remember it but I think I can figure it out:
Probability of Q failures is X^Q
Probability of Q1 failures is X^(Q1)*(1X)
Probability of Q2 failures is X^(Q2)*(1X)^2
etc.
So that generalizes to X^N*(1X)^(QN)
Use at your own risk though. There's a 42% chance I forgot everything I ever learned in statistics.

07302009, 12:00 PM



Location: Maryland not Murlin
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Quote:
Originally Posted by Evidenceiskey
if probability of failure is X
what is the probability of N failures in Q attempts?

Well, the probability of N failures in Q attempts is N/Q.
X is a fixed number, if I am reading your question properly, so it is a standard used to compare the result of N/Q. If the result is within 0.05 of X then it is a plausible outcome.
But, if you are asking for a comparison of N failures in relation to X, then it would be:
n!/(n1)!x!*p^x*q^nx
n=Q
x=QN
p=1X
q=X
Last edited by KLuv; 07302009 at 12:34 PM..
Reason: eggs!

07302009, 01:44 PM



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“The definition of insanity is doing the same thing over and over again and expecting different results”
Albert Einstein

07302009, 01:52 PM



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Quote:
Originally Posted by Evidenceiskey
if probability of failure is X
what is the probability of N failures in Q attempts?

It is the Binomial probablity distribution. Where failure = q and success = p.
Binomial probability  Wikipedia, the free encyclopedia
BUT this assumes that the events are INDEPENDENT.

07302009, 09:14 PM



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Oops... that's right. My equation only works for the probability of a specific order.
i.e. you have X^N*(1X)^(QN) chance of getting N failures and then QN successes.
Gotta get the Q!/(N!(QN)!) term in there.

07312009, 01:13 PM



Location: Maryland not Murlin
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Quote:
Originally Posted by pcity
Gotta get the Q!/(N!(QN)!) term in there.

This appears to be an nCr function, n!/(nr)!r!, which is used for counting the number of possible combinations of r and would yield a result of the total number of ways (combinations) in which a failure can happen.
Maybe that is what the OP is looking for? But I dunno, the guy hasn't posted a response yet. All I know is that there is a factorial in there somewhere.

07312009, 01:17 PM



Location: Maryland not Murlin
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Quote:
Originally Posted by hartford_renter
BUT this assumes that the events are INDEPENDENT.

Yes, a failure cannot affect the outcome of another failure if another failure where to occur. It would be nice if the OP could repost with the data....yes, I am a math nerd.

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