1. ## exponential distribution

A system with three independent components works correctly if at least one
component is functioning properly. Failure rates of the individual components are λ1
= 0.0001, λ2 = 0.0002, and λ3 = 0.0004 (assume exponential lifetime distributions).
(a) Determine the probability that the system will work for 1000 hours.
(b) Determine the density function of the lifetime X of the system.

2. This is almost identical to numerous other threads you have posted in the last few weeks. What exactly are you struggling with?

The CDF of the exponential distribution is:

$\displaystyle P(X\leq x)=1-e^{-\lambda x}$
so
$\displaystyle P(X>x)=e^{-\lambda x}$

So, the probability that the first component is alive after 1000 hours is $\displaystyle P(X_1>1000)=e^{-\lambda_1 *1000}$
similarly for X2, and X3 :

$\displaystyle P(X_2>1000)=e^{-\lambda_2 *1000}$
$\displaystyle P(X_3>1000)=e^{-\lambda_3 *1000}$

You want P(X1 and X2 and X3 alive after 1000 hours)
$\displaystyle =P(X_1 > 1000) * P(X_2 > 1000) * P(X_3) > 1000$
$\displaystyle =e^{-1000\lambda_1 }e^{-1000\lambda_2 }e^{-1000\lambda_3 }$
$\displaystyle =e^{-1000(\lambda_1 + \lambda_2 + \lambda_3) }$

Part b
You want to find F(K) = P(system fails by time K)
So
1-F(K) = P(system still alive at time K)

You can easily find the right hand side by generalising your answer to part a.

3. Originally Posted by SpringFan25
This is almost identical to numerous other threads you have posted in the last few weeks. What exactly are you struggling with?

The CDF of the exponential distribution is:

$\displaystyle P(X\leq x)=1-e^{-\lambda x}$
so
$\displaystyle P(X>x)=e^{-\lambda x}$

So, the probability that the first component is alive after 1000 hours is $\displaystyle P(X_1>1000)=e^{-\lambda_1 *1000}$
similarly for X2, and X3 :

$\displaystyle P(X_2>1000)=e^{-\lambda_2 *1000}$
$\displaystyle P(X_3>1000)=e^{-\lambda_3 *1000}$

You want P(X1 and X2 and X3 alive after 1000 hours)
For the first part you need the probability P(X1 or X2 or X3 is alive after 1000 hours)

4. true, i should read the question more

5. how can i determine the density function
part b

6. Originally Posted by ilikec
how can i determine the density function
part b

Look at part a) if instead of $\displaystyle 1000$ hours we do the calculation for $\displaystyle $$x hours to get \displaystyle$$ F(x)$ then the probability that the system fails between $\displaystyle x$ and $\displaystyle x+\delta x$ hours is $\displaystyle F(x)-F(x+\delta x)$ so in the limit the density of the life time is:

$\displaystyle f(x)=-\dfrac{d}{dx}F(x)$

CB

7. captain black isn't this a order statistic sum?

8. Originally Posted by ilikec
captain black isn't this a order statistic sum?
If you have generalised part a) it is irrelevant for part b). The generalised solution to part a) is the probability that the system has not failed by x hours. From that you can calculate the density without further reference to the earlier part of the question.

CB

9. Generalised solution to part a)

The probability that at least one sub-system is still working at time $\displaystyle $$t is 1 minus the probability that all have failed before \displaystyle$$ t$. This is:

$\displaystyle F(t)=1- (1-e^{-\lambda_1t})(1-e^{-\lambda_2t})(1-e^{-\lambda_3t})$

CB

10. captain black your generalized solution is for 1st order statistic. Right?

11. Originally Posted by ilikec
captain black your generalized solution is for 1st order statistic. Right?
Why are you concerned about the name rather than is it right? It is (if I have not made a mistake) the probability that the system is still operational at time t, which is the probability that one or more sub-systems is still functional at t.

Also I am not that familiar with order statistics and would have to look them up if I were to try to answer that question, which you can do as easily as me

CB