An engineering system problem

Question

An engineering system consisting of n components is said to be a k-out-of-n system (k ≤ n) if the system functions if and only if at least k of the n components function. Suppose that all components function independently of each other.

(a) If the ith component functions with probability Pi , i = 1, 2, 3, compute the probability that a 2-out-of-3 system functions.

(b) Let p1 = 0.5, p2 = 0.2, p3 = 0.1, compute the probability that a 1-out-of-3 system does not function.

(c) Compute the probability that a k-out-of-n system functions when all the pi equal p (that is, pi = p, i = 1, 2, ..., n).

Answer 1

In this problem, A~ is the complement of A

Let A be component 1, B be component 2, and C be component 3

Then, P(AB) = P1P2, P(AC) = P1P3, P(BC) = P2P3, P(ABC) = P1P2P3

P(2 or more components) = P(2 components) or P(3 components) =

P(ABC~) + P(AB~C) + P(A~BC) + P(ABC)

P(AB) = P(ABC) + P(ABC~), so P(ABC~) = P(AB) - P(ABC) = P1P2 - P1P2P3

P(AC) = P(ABC) + P(AB~C), so P(AB~C) = P(AC) - P(ABC) = P1P3 - P1P2P3

P(BC) = P(ABC) + P(A~BC), so P(A~BC) = P(BC) - P(ABC) = P2P3 - P1P2P3

Thus,

P(2 or more components) =

P(ABC~) + P(AB~C) + P(A~BC) + P(ABC) =

P1P2 - P1P2P3 + P1P3 - P1P2P3 + P2P3 - P1P2P3 + P1P2P3 =

P1P2 + P1P3 + P2P3 - 2 P1P2P3

b) A 1 out of 3 system does not function iff all 3 components fail =

(1 - P1)(1 - P2)(1 - P3) = (1 - .5)(1 - .2)(1 - .1) = .5 * .8 * .9 = .36

c) P(k out of n system functions) = P(k or more of n components function) =

∑_j=kⁿ C(n, j) p^j (1-p)^n-j (The sum means the sum from j = k to j=n; normally, we would write this with j=k at the bottom and j = n at the top)