Question

For a redundant 3 out of 5 parallel system, with A, B, C, D, and E being the five system components, how many common cause events can result in the failure of the system?

a) 3

b) 5

c) 10

d) 16

Answer 1

In a 3 out of 5 redundant parallel system, there are 10 common cause events that can result in the system's failure, calculated using the combination formula C(5, 3).

Step-by-step explanation:

The student's question pertains to a redundant parallel system, specifically how many common cause events can result in the failure of a 3 out of 5 system configuration. A redundant parallel system is designed to increase reliability by allowing multiple components to perform the same function; the system continues to operate even if some of the components fail. In a 3 out of 5 system, at least three components (A, B, C, D, E) must function for the system to work. When considering common cause events that can fail the system, we need to consider the combinations that can fail simultaneously.

To fail the system, at least 3 components need to be knocked out. Therefore, we are interested in the combinations of events that could cause the simultaneous failure of at least 3 components. The combination of any 3 failures can come from the 5 components, which can be calculated using the combination formula C(n, k) = n! / (k!(n - k)!), where n is the total number of components and k is the number of components that need to fail. Here, n = 5 and k = 3, so we have:

C(5, 3) = 5! / (3!(5 - 3)!) = (5 × 4) / (2 × 1) = 10

Thus, there are 10 common cause events that can result in the failure of the redundant parallel system.