A basic probability question

[Ford Prefect]

Hello everyone. I'm having some trouble with some mathematics. Though it is hilariously simple compared to some of the stuff that gets bandied about here, I felt that if anyone could explain this to me, it would be someone on SDN.

It's a Poisson problem, and while the first couple of parts were simple enough, I'm having trouble wrapping my head around a certain question. It is basically asking about the probability of faults in a optic cable. The following is pertinent information:

"The rate at which faults occur on optical fibre cable is believed to follow a Poisson distribution with an average of µ = 0.0000092 (i.e. 9.2 * 10-6) faults per kilometre per year.

The defence department reliability specification requires that there is no more than a 5% probability of the cable having a fault over a ten year period."

The question I'm having trouble with is as follows:

"If TransTel build both the shortest (2800km) and second shortest (5400km) paths, determine the probability that at least one path has no faults over a ten year period. Explain whether this configuration meets the reliability specification."

My initial problem was that I was treating it as one big long path, which was rather silly. However, I just can't seem to fathom out the correct way to go about this problem. If anyone would care to give me some pointers, that would be wonderful.

[dragon]

Here's a good site for this link, they have inside a link to a poisson calculator that can calculate failure rate, time frame, probability and more such. Other than that its been almost 10 years since I've done probability so can't help to much.

[WacoKid]

Wouldn't you just multiply the lengths individually times the probablility to get the failure rate?

And then do P1(fail)P2(not fail)+(P1(not fail)(P2(fail)+(P1(not fail)(P2(not fail)?

Been a while since I've done probability too, just a guess.

[fnord]

Well, the first thing that leaps out at me is that, since faults appear to be independent of each other, is to first determine the probability of each path getting through the ten year period with no faults, then, taking Pi as such probability,

Psystem = 1 - (1 - Ps) * (1 - Pl)

Where Ps, Pl and Psystem are the chances of the short path, long path, and whole cable system respectively, getting through the ten years without developing a fault.

From here on in, I'm probably about as in the dark as you are.

[fnord]

Yeah, might as well follow this through.

Per yon PDF,Poisson of independent sum is a poisson with parameter equal to sum of component parameters.

Poisson parameters are then (as well as expected number of faults):
0.2576
and
0.4968

Probability of k failures with parameter l
(l^k * exp(-l)) / k!

Thus, probability of zero failures simplifies to
exp(-0.2756) = 0.772904332
and
exp(-0.4968) = 0.608474666
respectively (to 9 dec pl, on a casio fx-82tl calculator)

Psystem then becomes
1 - (1 - 0.772094332) * (1 - 0.608474666)
0.911 or so - namely, system has roughly 91% chance of getting at least one path through the ten year period without fault developing, which is worse than the DoD requirement in the question.

[Kuroneko]

The 2800km cable has an expected number of faults λ₁ = 0.02576 per year, while the 5400km cable has λ₂ = 0.04968 faults per year. The probability that there are zero faults on a given cable in a year is λ⁰exp(-λ)/0! = exp(-λ), so that the probability that at least one cable remains functional in a year is the complement of both failing:
[1] P = 1 - [1-exp(-λ₁)][1-exp(-λ₂)] = 0.9988.

[Ford Prefect]

Aha, thank you all. I have now got an idea of how to do this question properly.

[dragon]

Aha, thank you all. I have now got an idea of how to do this question properly.

I'm glad someone does as they more than lost me, and the sad thing is I have a degree in mathematics But thats what I get for not praticing this stuff for almost 10 years.

[Ford Prefect]

This is basically the first time that Kuroneko has posted some mathematics and I was able to understand. I'm revelling in it.

[fnord]

Kuroneko,

How come you worked out the probability of failure over a year, then combined them to produce a system failure probability, compared to my approach? What did I botch up?

[Kuroneko]

Well, this is fairly embarrassing. I did λ = [9.2E-6/(yr·km)][2800km][1 yr] = 0.02576, etc., whereas the DoD criterion was over 10 years, rather than 1 year. You're completely right, fnord, and apparently I can't read.

[Ford Prefect]

I hate questions which have answers which don't fit stated requirements. I was hoping that it would, but there is a subsequent question which makes use of a third path, which may in fact fit.

[fnord]

Go ahead, post said subsequent question - you've managed to get me all interested.

[Ford Prefect]

I have already completed. However, I have a normal distribution question I can't work out.

The mean is 3.05, the standard deviation is 0.025. The question is asking about ingots, which have to be between 2.97 and 3.08 in order to be in-spec. While I know the probability of a randomly selected ingot being in-spec is 0.8842. However, the question is asking what is the probability of 48 out of a run of 50 being in-spec, and I have no idea how I did it, and I lost my original workings.

[Kuroneko]

The mean is 3.05, the standard deviation is 0.025. The question is asking about ingots, which have to be between 2.97 and 3.08 in order to be in-spec. While I know the probability of a randomly selected ingot being in-spec is 0.8842.

Yes.

However, the question is asking what is the probability of 48 out of a run of 50 being in-spec, and I have no idea how I did it, and I lost my original workings.

Hints (assume independence):
1. What is the probability that only the first two ingots are not in-spec? Probability of only ingots #42 and #44 being bad? Does it matter which two I specify?
2. How many pairs could I specify, anyway?
Your final answer should be about 0.045.

[Ford Prefect]

I apologise Kuroneko, but I honestly cannot fathom it out. I have not concerned myself with what ingot is out of spec, simply because I haven't been able to get that far.

[Kuroneko]

The probability of being in-spec, is, as you've derived, p = 0.8842. The probability of being not in-spec is then just 1-p. Therefore, the probability that only the first two are failures is
P(I1,I2) = P(I1 out-spec)P(I2 out-spec)P(I3 in-spec)P(I4 in-spec)...P(I50 in-spec) = p^48(1-p)^2.
If I repeated this with, say, only the third and fifth being out-spec, would right-hand side change? If not, then you're just adding up the same number for each pair of ingots that can fail, i.e., multiplying it by the number you'd get for hint #2 above.

[fnord]

Another hint: the normal distribution stuff is a large, juicy, red herring.

[Ford Prefect]

Oh, that makes sense! Thank you very much. Now all that remains is a confidence interval question.