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) =
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).