# Thread: Probability of a Compenent failing

1. ## Probability of a Component failing

A critical component on a submarine has an operating lifetime that is exponentially distributed with a mean of 6 months. As soon as a component fails, it is replaced bya new one having statistically identical properties. What is the minimum number ofspare components which the submarine ought to carry if it is leaving for a one yeartour of duty and it is desired that the probability of having an inoperable unit causedby failures exceeding the spare inventory be less than 0.02

Let V represent the critical component

$\displaystyle E(V) = 0.5$ =>$\displaystyle \lambda = 2$

The probability of the first time the component fails is:

$\displaystyle P(V \ fails) = 2e^{-2}$

Kind of stuck here. Any hints/suggestions would be greately appreciated.

2. How about this: Suppose $\displaystyle X_1, ..., X_n$ are iid and distributed according to $\displaystyle \mbox{Exp}(1)$. Then

$\displaystyle \displaystyle \sum_{i = 1} ^ n X_i \sim \mbox{Gamma}(n, 1)$.

3. So $\displaystyle \displaystyle \sum_{i = 1} ^ n V_i \sim \mbox{Gamma}(n, 2)$

EDIT: Okay, I think I got it

I need to find the probability that

So $\displaystyle \displaystyle P(\sum_{i = 1} ^ n X_i <1)<0.02$

$\displaystyle E(\Sigma X_i) = 0.5n$ $\displaystyle Var(\Sigma X_i)=0.25n$

Applying the central limit theorem:

$\displaystyle P(Z<\frac{1-0.5n}{0.25\sqrt{n}})$<0.02

$\displaystyle \frac{1-0.5n}{0.25\sqrt{n}}=2.05$

Solving for n, I get n=0.9835

So they need to bring 2 components.

Have I made a mistake anywhere?

4. I'm not as bold as some, but I'm pretty sure using the CLT for n = 2 isn't a good idea.

Gammas are related to chisquare random variables, so you have tables that give you the quantiles for them.

5. Originally Posted by theodds
I'm not as bold as some, but I'm pretty sure using the CLT for n = 2 isn't a good idea
I'm not sure what you meant by that. I didn't use the CLT for n = 2. I used to to solve for n and got n =2

6. But the CLT isn't, typically, a good approximation for n = 2. The reasoning behind what you did only applies when the $\displaystyle n$ that you get is sufficiently large the the CLT to hold.

(1) The n I need is large, so apply the CLT
(2) Solve for n
(3) Get n = 2

You see the problem, yes?

7. Originally Posted by theodds
Gammas are related to chisquare random variables, so you have tables that give you the quantiles for them.
A Chi-Square is a special case of the gamma where $\displaystyle \Gamma (n,2)$~Chi-Square where n is the degrees of freedom. But I still don't know what the degrees of freedom are.

8. Originally Posted by statmajor
A Chi-Square is a special case of the gamma where $\displaystyle \Gamma (n,2)$~Chi-Square where n is the degrees of freedom. But I still don't know what the degrees of freedom are.
Actually, it is the same as a chisquare with 2n degrees of freedom (not n).

You are supposed to choose n, or equivalently, the degrees of freedom of an appropriate chi-square. Here's one more hint:

$\displaystyle \displaystyle Z \sim \mbox{Gamma}(\alpha, \beta) \iff \frac{2Z} \beta \sim \chi^2 _{2\alpha}$
where we are using the shape/scale parameterization of the Gamma (i.e. $\displaystyle \mbox E Z = \alpha\beta$). You can look up in a table an appropriate cutoff to get the n you want.