Probability of failure.............. (function, nerd, comparison)

[Evidence-is-key]

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

[GregW]

How about asking how long until the first failure?

[Evidence-is-key]

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?

[pcity]

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.

[K-Luv]

Originally Posted by Evidence-is-key

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

[Asheville Native]

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

Albert Einstein

[hartford_renter]

Originally Posted by Evidence-is-key

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.

[pcity]

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.

[K-Luv]

Originally Posted by pcity

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.

[K-Luv]

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.