SOLUTION: An electronic assembly consists of 2 sub systems A and B. from the previous testing procedures the following probabilities are known:
P(A fails)=0.2
P(B fails alone)=0.15
P(A an
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-> SOLUTION: An electronic assembly consists of 2 sub systems A and B. from the previous testing procedures the following probabilities are known:
P(A fails)=0.2
P(B fails alone)=0.15
P(A an
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Question 118388: An electronic assembly consists of 2 sub systems A and B. from the previous testing procedures the following probabilities are known:
P(A fails)=0.2
P(B fails alone)=0.15
P(A and B fails)=0.15
evaluate P(A fails given B fails) {conditional probability}
and
P(A fails alone). Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! P(A fails)=0.2
P(B fails alone)=0.15
P(A and B fails)=0.15
evaluate P(A fails given B fails) ------------------
and
P(A fails alone).
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An electronic assembly consists of 2 sub systems A and B. from the previous testing procedures the following probabilities are known:
P(A fails)=0.2
P(B fails alone)=0.15
P(A and B fails)=0.15
evaluate P(A fails given B fails) {conditional probability}
and
P(A fails alone).
-------------------
P(A fails | B fails) = [P(A fails and B fails)]/[P(B fails] = 0.15/P(B fails)
but P(B fails) = P(B fails alone) + P(B fails and A fails) = 0.15 + 0.15 = 0.30
So P(A fails|B fails) = 0.15/0.30 = 1/2
-------------------------
P(A fails alone) = P(A fails and B does not)
=P(B does not fail|A fails)*P(A fails)
=(1-P(A and B fails))*0.2
=(1-0.15)*0.2
=(0.85)*0.2
= 0.17
=================
Cheers,
Stan H.
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