P(A to fail) =0.1, P(B to fail) =.05; The system fails if any of the two A or B fails

Is the solution method correct given below ?
P(A to fail) =0.1
P(B to fail) =.05
The system fails if any of the two A or B fails,, find the probability that A has failed when system fails given that only one component has failed ??

P(A') + P(B') =1 since system has failed and only one component must fail
P(B') = 1/2 * P(A')
=> P(A') + 1/2 * P(A') =1
=> P(A') = 2/3

Please help as i have done this in the exam,, i know the answer is correct but instructor has given me zero marks so as an evidence i must show this to him ......

Syed Mansoor said:

Is the solution method correct given below ?
P(A to fail) =0.1
P(B to fail) =.05
The system fails if any of the two A or B fails,, find the probability that A has failed when system fails given that only one component has failed ??

P(A') + P(B') =1 since system has failed and only one component must fail
P(B') = 1/2 * P(A')
=> P(A') + 1/2 * P(A') =1
=> P(A') = 2/3

Please help as i have done this in the exam,, i know the answer is correct but instructor has given me zero marks so as an evidence i must show this to him ......

Click to expand...

You are correct in that you must have relationship between
P₁ = {p~a=P(A will fail)=0.10, pa=P(A will not fail)=0.90}
and
P₂ = {p~b=P(B will fail)=0.05, pb=P(B will not fail)=0.95}.
However, the problem as stated does not give that relationship. If we assume that P₁ and P₂ are independent, then we can set the problem up as: Take a 1000 cases and let
-n1=number of time we had A fail and B fail = 1000 * (p~a * p~b) = 5
-n2=number of time we had A not fail and B fail = 1000 * (pa * p~b) = 45
-n3=number of time we had A not fail and B not fail = 1000 * (pa * pb) = 865
-n4=number of time we had A fail and B not fail = 1000 * (p~a * pb) = 95

What formula, which you should remember for a test (or work it out, but that takes more time), does this suggest to you? What do you think the answer is assuming P₁ and P₂ are independent?

Syed Mansoor said:

Is the solution method correct given below ?
P(A to fail) =0.1 & P(B to fail) =.05
The system fails if any of the two A or B fails,, find the probability that A has failed when system fails given that only one component has failed ??
P(A') + P(B') =1 since system has failed and only one component must fail
P(B') = 1/2 * P(A')
=> P(A') + 1/2 * P(A') =1
=> P(A') = 2/3

I know the answer is correct but instructor has given me zero marks so as an evidence i must show this to him ......

Click to expand...

There is absolutely no way that either of those two is necessary true. Why do you think so?
As presented the question has no answer. Have you left out some other bit of necessary information?
I think that your notion may be the source of confusion.
Let \(\displaystyle \bf{A}\) mean that component A fails. So \(\displaystyle \bf{A'}\) means it does not fail.

The question "find the probability that A has failed when system fails given that only one component has failed" means:
\(\displaystyle \mathcal{P}\left( {\bf{A}|(\bf{A} \cap \bf{B'}) \cup (\bf{A'} \cap \bf{B})} \right)\)
Are you given that the failures are independent? Are you given the probability that the system fails?