Help me solve a probability question

[claudiocamera]

probability of two failures 100th 99th

I have doubt about the way to solving the following problem:

A system has 100 components. The probability that a specific component will fail in the interval equals exp(-a/T ) - exp(-b/T) . Find the probability that in the interval ( 0, T/4) no more than 100 components will fail.

Somebody helps?

 

[main_road]

Re: Probability Question

You have only 100 components. If everthing fails at the worst case scenario, you still have no more than 100 failed components. Hence, the probability is 1.

 

[claudiocamera]

Probability Question

The answer is just the oposite P aproximately zero, that is the challenge... No more tha 100 means that can fail 1,2,3....100, I stiil didnt get how to reach the answer...

 

[neils_arm_strong]

Re: Probability Question

since we need P(x<100),

P(x<100)=∫(exp(-a/T)-exp(-b/T)dt with t ranging from 0 to T/4.
Correct me if i am wrong.

 

[steve10]

Re: Probability Question

This can be solved by the binomial distribution.

Set N=100, and p=1-exp(-1/4) the probability for one component to fail in ( 0, T/4), while q=1-p=exp(-1/4) not to fail. Besides, set
C(n,m)=m!(n-m)!/n! --- combinatorial numbers (n>=m)

Then,we have the probabilities in all cases:

0 failure: C(N,0)q^N
1 failure: C(N,1)p*q^(N-1)
2 failures: C(N,2)p^2*q^(N-2)
3 failures: C(N,3)p^3*q^(N-3)
....
N-1 failures: C(N,N-1)p^(N-1)*q
N failures: C(N,N)p^N

Notice that C(N,0)=C(N,N)=1.

Based on those values, you can find the probabilities in more complicated cases. For example, the probability of exactly 1 or 3 failures is
C(N,1)p*q^(N-1)+C(N,3)p^3*q^(N-3).

 

[Nazgul]

Probability Question

Main road is right, you can't possibly have more than 100 components fail becuase your system is composed of only 100 components. Thus the probability of less than 100 components failing is 1.

You can also ask yourself this question, what is the probability of more than 100 components failing? The answer is zero. Then P(x<=100) = 1 - P(x>100) = 1

 

[claudiocamera]

Probability Question

Thank you all, but I think got the answer myself:

With a=0 T=1/4 we have :

p= 1- exp(-1/4)=0.22
np=220
npq =171.6
K=100
Using dmoivre laplace aproximation to the normal distribution we have :
(k-np)/(npq)^0.5 = - 9.16
hence: P{ 0<K<100} = G(-9.16) which is aproximately zero. It match with the answers given to the problem , I am almost sure it is by this way...

 

[steve10]

Re: Probability Question

1. n=1000?
2. When you set K=100, what is P{ 0<K<100}?

 

[claudiocamera]

Probability Question

Ok, let's see a review:
n=100
np=22
npq=17.16
k2=100
k1=0
(k2-np)/(npq)^0.5 = 18.82
(k1-np)/(npq)^0.5 = - 5.31

hence: P{ 0<K<100} = G(-18.82) - G(-5.31) wich is approximately 1 and is according to the answers given by the guys.

Well, according to this I come back to the first...I suppose the answer given in the book is either incorrect or the number of components is 1000.

 

[steve10]

Re: Probability Question

Let me guess ... according to P{0<K<100}, you wanted to calculate the probability of one failure, two failures, ... up to 99 failures. The result is the

C(100,1)p^1*q^99+C(100,2)p^2*q^98+...+C(100,99)p^99*q^1

(By the way, I made a mistake in defining C(n,m) in the previous post. It should be upside down, C(n,m)=n!/(m!(n-m)!), sorry)

This sum is very close to 1. Besides, this sum is considered very complex and should be replaced by a simpler formula. That is where the "dmoivre laplace" fits in. Accoding to this theorem, the sum above can be approximated by an integral with a normal density, which is "G(18.82) - G(-5.31)" (by the way, I changed a sign before 18.82).