In an RBD, how do you compute system reliability for a parallel arrangement of three identical components with reliability R?

• A. Rsys = R^3
• B. Rsys = 1 − (1 − R)^3
• C. Rsys = 3R − 2R^2
• D. Rsys = 1 − R

Answer: (B)

In a parallel arrangement, the system stays up as long as at least one component is functioning. The only way the system fails is if all components fail. With three identical, independent components each having reliability R, the chance a single component fails is 1 − R, and all three fail is (1 − R)^3. Therefore, the system reliability is 1 − (1 − R)^3. Expanding this gives 3R − 3R^2 + R^3, which shows why options like R^3 or 1 − R don’t fit a parallel three-component setup. This expression also behaves sensibly at the extremes: if R = 0, the system fails; if R = 1, the system is perfect.