# Simple Probability Statistics - Assembly Failure - Binomial Equations

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#### M Greenaway

Simple Probability Stats

OK Guys

If I have six components on an assembly, and the components come from a source that is 14% defective, how many of my assemblies will fail if one defective component causes the assembly to fail ?

Is it simply 0.14 + 0.14 + 0.14 + 0.14 + 0.14 + 0.14 = 0.84 or 84%

R

#### Rob Nix

Will the jury please disregard my last comment.

This is a binomial equation. n=6 trials, r=1 failure, from lot of 14% bad = 80% chance of at least 1 failure.

This is provided the source is large enough not to be affected much by 6 parts being removed.

Sorry I first reacted without brain in gear.

#### Mike S.

##### Happy to be Alive
Trusted Information Resource

#### Tim Folkerts

Trusted Information Resource
M Greenaway said:
If I have six components on an assembly, and the components come from a source that is 14% defective, how many of my assemblies will fail if one defective component causes the assembly to fail ?

It this case, think of how likely it is to work.

If there was just 1 component, then the odds of working are 1-0.14 = 0.86
With two components, the first has to work and the second has to work. The odds of both working are (0.86)*(0.86) = 0.740.
With 6 components, it becomes (0.86)^6 = 0.405.

So, 40.5% would work, and 49.5% would fail.

Tim

#### Claes Gefvenberg

I agree Tim,

...and welcome to the Cove

/Claes

M

#### M Greenaway

If 40.5% work then 59.5% must fail.

But what is the probability that an assembly will have one or more defective components from a pool of 14% defective components ? Is it the same ??

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#### Rob Nix

My original reply (which I deleted) was based on what Tim said (only in reverse - that was my error). But then I realized that the equation Tim is using is for system (un)reliability, where 86% is a unit's reliability. But Greenaway's question (and the second question) is about a discrete population of parts, which lend credence to the binomial distribution, which is "if the probability of occurrence p is constant on each of n independent trials, then the probability of r occurrences in n trials is: n!/(r!(n-r)!)pq^-r". To simplify matters I just used Excel's "binomdist" formula.

If I am way off base here, please explain.

M

#### M Greenaway

I am now thinking that it is of one or more components is defective then the assembly fails, hence we need to OR all the possible combinations of ANDs.

So the assembly fails if we have one defective component, or two, or three, or four, or five, or six (all).

Which would be 0.14 + 0.14^2 + 0.14^3 + 0.14^4 + 0.14^5 + 0.14^6

This questioning relates to the fact that in reality I have 52% defective assemblies from a pool which appears to be 14% defective components (6 components per assembly).

Hence I want to know is it random throughout the batch, or do I have mixed batches, one all good, and one (or more) with random defects.

Any ideas ??

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#### Rob Nix

The probability of zero or one defect is 80% ( arghh - this is what I get when I try to do three jobs at once). The probability of no defects is 40.5% (like the others said, only using a different method), therefore the probability of one or more is 59.5%. You had 52%. Using control limits for a p-chart, and if your sample size (for the number of assemblies you checked to arrive at 52%) was less than about 350, then you may assume that it is random. NOTE: Do you have p-chart data on these assemblies? The patterns in the chart may give an indication of your other possible root causes.

#### Tim Folkerts

Trusted Information Resource
M Greenaway said:
I am now thinking that it is of one or more components is defective then the assembly fails, hence we need to OR all the possible combinations of ANDs.

So the assembly fails if we have one defective component, or two, or three, or four, or five, or six (all).

Which would be 0.14 + 0.14^2 + 0.14^3 + 0.14^4 + 0.14^5 + 0.14^6

Right idea, but not quite complete.

The odds of all six being bad are indeed
0.14^6
The odds of one being good and five bad are
(0.86) * (0.14^5) * 6.
(The 0.86 is the odds that the first is good; the 0.14^5 are the odds that the next five are all bad; the 6 is for 6 choices for which one is bad.)

The odds of two good and four bad are
(0.86)^2 * (0.14)^4 * (6*5/2)
The odds of three good and three bad are
(0.86)^3 * (0.14)^3 * (6*5*4/2/3)
The odds of four good and two bad are
(0.86)^4 * (0.14)^2 * (6*5*4*3/2/3/4)
The odds of five good and one bad are
(0.86)^5 * (0.14)^1 * (6*5*4*3*2/2/3/4/5)

Then, as you say, we need to add up the five terms. Plug it into excel and you get a total of 59.5% bad, which is what I got before .

NOTE: using the same formula, the odds of all sixe being good are
(0.86)^6 * (0.14)^0 * (6*5*4*3*2*1/2/3/4/5/6)
= (0.86)^6 *1 * 1
(again in agreement with my previous calculations).

Tim

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