Question 629744: There are three components that operate in a series so all must function properly if it is to be complete. The probability of failure during one period of operation are .03, .05, .06. Component failures are statistically independent.
a) Fine the probability that the system functions properly.
b) Find the probability that the system fails during one period of operation
This is what I have done so far but I'm not sure its the right answer please help me and show how you got to the answer so I understand
a) .97*.95*.94=.866
b)(.03+.05+.06)-(.03*.05*.06)=.14
Found 2 solutions by Theo, ikleyn:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i don't have time to solve it but this is what you have to do.
assume the components are a, b, c
probability of failure are .03, .05, .06
probability of a good system is .97 * .95 * .94
you are correct there.
probability that the system fails during one part of the operation is 1 - the probability that the system doesn't fail.
that's the easy answer and it's equal to 1 - .86621 = .13379
the hard answer is that the probability that the system will fail is equal to:
p(a fails) + p(b fails) + p(c fails) - p(a and b fail at the same time) - p(a and c fail at the same time) - p(b and c fail at the same time + p(a and b and c fail at the same time).
that's the kicker.
that last one.
it has to be added.
no time to explain but if you go through the numbers you will see that the probability of the system failing equals .13379 doing it the hard way.
write if i'm wrong or if you have questions.
that's all the time i have for now.
Answer by ikleyn(53751)  (Show Source):
You can put this solution on YOUR website! .
There are three components that operate in a series so all must function properly if it is to be complete.
The probability of failure during one period of operation are .03, .05, .06.
Component failures are statistically independent.
a) Fine the probability that the system functions properly.
b) Find the probability that the system fails during one period of operation
This is what I have done so far but I'm not sure its the right answer please help me and show
how you got to the answer so I understand
a) .97*.95*.94=.866
b)(.03+.05+.06)-(.03*.05*.06)=.14
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I will explain it in a short and clear way.
(a) The system functions properly, if and only if all three components function properly.
Therefore, since the components are statistically independent,
P(the system functions properly) = (1-0.03)*(1-0.05)*(1-0.06) = 0.97*0.95*0.94 = 0.86621.
(b) P(the system fails) = 1 - P(the system works properly) = 1 - 0.86621 = 0.13379.
The probability to fail is the COMPLEMENT to the probability to function properly.
It is the MAJOR CONCEPTION to learn from part (b) of this problem.
Solved.
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