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" .
The correct answer should be:
P(A → B) = p { 1 − (1 − p) {1 − [1 − (1 − p)^3] p }} [1 − (1 − p)^2]
great demonstration for the devide and conquer method :)
Very helpful. The problem's difficulty was instructive. Thank you very much!
Thank you Sir. Your explanation was laconic.
thanks Prof.Kuang Xu
in 7:05, i guess the last term p^2 should be only p
i think that square is actually the upper part of closing second bracket
how about if the system is not series nor parallel? ie like complex one
I think the final answer is incorrect.
Does anybody know why the probability of failure under a parallel system is not the sum of 1-p?
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" .