Probability of Defects - A Problem

W

wallacemd

#1
Probability of defects ?

Here is a question for you CQE's out there.

The probability that defect type A occurs is 0.83, the probability that defect type B ocurs is 0.83, and the probability that both occur is 0.83. Find the probability that at least one of the defects occurs.

The answer is supposed to be 0.029 but I am a baffled as to how this is achieved.

I appreciate any help.:bigwave:
 
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Marc

Hunkered Down for the Duration
Staff member
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#3
I'm not the person to answer, but there was a good discussion of probability in a recent thread involving Inspection - the 80-20 Rule in particular.

I think it's add the probabilities and divide by 3 - heck - I'd have to dig out my stats book - and I'm too lazy and might mis-interpret.

That said, maybe Rick Goodson can lend a hand here.
 
W

wallacemd

#4
I am assuming it is a single part.

The problem originates from the CQE Handbook supplemental section published by ASQ press. It appears to be a fundamental probability problem requiring application of the general addition rule but the probabilities do not seem logical. Perhaps the question is in error.

Comments are welcome.

:bonk:
 
D

D.Scott

#5
Here I go again - always sticking my 2 cents in.

Anyway, I am not a statistician but here goes.

I think somebody messed up the question. The probability of both occuring is superfluous. The question relates to a single occurance of either instance. The chance of either one is obviously .83 on any pick for either A or B. As A+B could never happen without success on one of the first two conditions, the question must revert to "what is the possibility of either event happening" (.83 + .83) - (.83*.83) = .97

If you were to ask what is the chance of two good consecutive samples with no defects, you would use the chance of a good sample on the first try (.17) times the chance on the second part (.17) or, 17*.17 = .0289 (.029).

Then again, what do I know? Where is Kevin when you need him? LOL

Dave
 
R

Rick Goodson

#6
Sorry I didn't responded earlier, I have been on the road and out of touch for a while.

First, this is one of those dumb questions, which I really hate, that the certification committees develop to confuse people. The answer usually lies in understanding the intent of the arcane question. Never the less we will forge ahead. Second, IMHO our esteemed colleague Mr. Scott has hit the nail on the head as usual. As he stated, I believe there is an error in the question.

By definition in the problem, the probability of defect A does NOT preclude the probability of defect B, therefore the events are not mutually exclusive. Given they are not mutually exclusive then the probability of independent event A or independent event B or both is the probability of A plus the probability of B minus the probablity of both:

P(A or B or both) = P(A) + P(B) - P(both)

Example, what is the probability of obtaining a jack or a diamond on one draw from a deck of cards. The probability of a jack is 4/52 = 0.077. The probability of diamond is 13/52 = 0.327. The probability of the jack of diamonds is 1/52 = 0.019. Therefore the answer is 0.077 + 0.327 - 0.019 = 0.308.

So, in the case in point the probability of both occurring is the probability of A, 0.83, multiplied by the probability of B, 0.83, which equals 0.6889. The probability of A or B or both is 0.83 plus 0.83 minus 0.6889 which is 0.9711.

If the probability of defect A or B or both is 0.9711 then the probability of no defect is 1.0000 - 0.9711 = 0.0289 or 0.029.

However, if we accept the probabilities given in the question (0.83 - 0.83 + 0.83 = 0.83) then the probability of no defects is 1.00 - 0.83 = 0.17 and as Mr. Scott states, the answer of 0.029 is the probability of no defects in two consecutive samples.

Regards,

Rick
 
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