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Old 07-30-2009, 08:03 AM
 
434 posts, read 854,236 times
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if probability of failure is X
what is the probability of N failures in Q attempts?
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Old 07-30-2009, 08:35 AM
 
Location: Londonderry, NH
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How about asking how long until the first failure?
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Old 07-30-2009, 08:40 AM
 
434 posts, read 854,236 times
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Quote:
Originally Posted by GregW View Post
How about asking how long until the first failure?
???

i need to know what the probability of N failures in Q attempts?
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Old 07-30-2009, 10:49 AM
 
3,137 posts, read 7,896,688 times
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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 Q-1 failures is X^(Q-1)*(1-X)
Probability of Q-2 failures is X^(Q-2)*(1-X)^2
etc.

So that generalizes to X^N*(1-X)^(Q-N)

Use at your own risk though. There's a 42% chance I forgot everything I ever learned in statistics.
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Old 07-30-2009, 12:00 PM
 
Location: Maryland not Murlin
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Quote:
Originally Posted by Evidence-is-key View Post
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!/(n-1)!x!*p^x*q^n-x

n=Q
x=Q-N
p=1-X
q=X

Last edited by K-Luv; 07-30-2009 at 12:34 PM.. Reason: eggs!
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Old 07-30-2009, 01:44 PM
 
16,301 posts, read 24,263,079 times
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“The definition of insanity is doing the same thing over and over again and expecting different results”

Albert Einstein
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Old 07-30-2009, 01:52 PM
 
1,679 posts, read 2,490,271 times
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Quote:
Originally Posted by Evidence-is-key View Post
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.
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Old 07-30-2009, 09:14 PM
 
3,137 posts, read 7,896,688 times
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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*(1-X)^(Q-N) chance of getting N failures and then Q-N successes.

Gotta get the Q!/(N!(Q-N)!) term in there.
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Old 07-31-2009, 01:13 PM
 
Location: Maryland not Murlin
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Quote:
Originally Posted by pcity View Post

Gotta get the Q!/(N!(Q-N)!) term in there.
This appears to be an nCr function, n!/(n-r)!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.
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Old 07-31-2009, 01:17 PM
 
Location: Maryland not Murlin
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Quote:
Originally Posted by hartford_renter View Post
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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