If I have 4 possible states (A,B,C,D) that can lead to a system state F by the following relation (A OR B) AND (C OR D) = F how would I find the probability of the final failure state happening?
I remember doing this in school, but I’ll be damned if I can remember how to do it now.
You need to know the probabilities of A, B, C and D, including whether or not they are independent. Assuming they are independent, the probability of (A or B) is
P(A) + P(B) - P(A)P(B)
and the probability of (C or D) is
P© + P(D) - P©P(D)
and the overall probability is the product of these two.
he didn’t say exlusive OR
I think it should be
( P(A) + P(B) ).( P© + P(D) )
In addition to what Jabba posted (I totally agree), the probability of (A or B) generally is
P(A) + P(B) - P(A and B), where
P(A and B) = P(A|B) * P(B) = P(B|A) * P(A)
If they are independent, P(A|B) = P(A) and P(B|A) = P(B), which gives you the formulas as seen in the previous post.
Neither Jabba nor myself assumed an exclusive OR.
Subtracting P(A and B) merely makes sure that this probability isn’t counted twice (the intersection is contained both in P(A) and in P(B)).
whoops, you are right MartinL & Jabba, please disregard my post.
Serves me right for jumping in… sorry!
Thanks folks, helps a lot.
If either (A or B) simultaneous with (C or D) can lead to the state F, then A and B are counted as parallel, as are C and D.
The probability (A or B) is (A + (1-A)(B)).* So if A can happen 90% of the time, and B can also happen 90% of the time, you have
.90 (chance that A can happen) +
.10 (chance that A doesn’t happen) * .90 (chance that B will happen when A fails) =
.99
The probability of the system achieving F on any one trial is
(A + (1-A)(B)) * (C + (1-C)(D)).
It makes it easier to think of A, B, C, and D as “black boxes” that are chained together. A and B share an input and output cable, as do C and D. C-D’s input cable is directly connected to A-B’s output. Each box has a known probability of successfully passing the signal.
Now you can simply think of this as a system reliability problem. If you send a signal into the ABCD box, what are the odds that you’ll get the same signal back out?
If the A and B are both .90 reliable and C and D are both .80 reliable, then R(ab) = .99 and R(cd) = .96, and R(abcd) = .9506 .
You’ve combined four components, all with reliability at or below 90%, and built a machine that is (only just) better than 95%.
(by the way, I didn’t realize it, but Jabba’s post contains the same formula as mine in a slightly different form).
Enjoy,
Jurph