Question

P(A) = 0.78
P(B) = 0.78
P(C) = 0.74
P(D) = 0.74

What's the probability of this system working?

Answer 1

not sure what this means exactly but you might try \[\frac{1}{3}\times \left(.78+.78+.74^2\right)\]

Answer 2

depends on what the probability of entering each path is i think

Answer 3

is P(A) the reliability of A?

Answer 4

yeah P(A) is probability that A works.

I need one of the lines to work in order for the whole thing to work

Answer 5

if so you're looking at
P(working)=1-P(failure)=1-(1-P(a))(1-P(B))(1-P(C or D))
=1-(1-P(a))(1-P(B))(1-P(C)P(D))

Answer 6

wow that's a great way of thinking, thanks nuts.

Answer 7

wondering why \[1-(1-P(A))\] rather then just \[P(A)\]

Answer 8

well it's 1-P(all paths failing)
=1-P(path 1 fails)P(path 2 fails)P(path 3 fails)
=1-(1-P(path 1 works)(1-P(path 2 works)(1-P(path 3 works))
and path 3 comprises component C and D so you must take the product of the reliabilities of C and D

Answer 9

@satellite73