Network Reliability

[@shinzhao6978]

The correct answer should be: P(A → B) = p { 1 − (1 − p) {1 − [1 − (1 − p)^3] p }} [1 − (1 − p)^2]

[@MinhNguyen-gg2iu]

great demonstration for the devide and conquer method :)

[@mayteylg]

Very helpful. The problem's difficulty was instructive. Thank you very much!

[@khalidsaifullah1321]

Thank you Sir. Your explanation was laconic.

[@eren96lmn]

in 7:05, i guess the last term p^2 should be only p

[@newb_embedded040]

i think that square is actually the upper part of closing second bracket

[@ockeyzhu1688]

thanks Prof.Kuang Xu

[@kreigu4700]

how about if the system is not series nor parallel? ie like complex one

[@SH-jq2vq]

Does anybody know why the probability of failure under a parallel system is not the sum of 1-p?

[@gabrieldjebbar5057]

Well, its basically the same as the case of the coin toss experiment ! If the probability of having a head is (1-p) , then probability of having 3 heads in a row (or at the same time) is (1-p)^3 ... by anology here we get the same result changing "tail" into "no failure" and "head" into "failure" .

[@2aa22a]

I think the final answer is incorrect.