A system made up of 5 series components has a design reliability of .96. Subsequently, 4 of the components have reliabilities apportioned to them of 0.992, 0.991, 0.994, and 0.992. What should the…

Question

Options

• A0.970
• B0.969
• C0.990
• D0.992

• A
  16% (8)
• B
  6% (3)
• C
  74% (37)
• D
  4% (2)

Explanation

For a series system, the system reliability equals the product of all component reliabilities: R_sys = R1 × R2 × R3 × R4 × R5. To find R5, divide the system reliability by the product of the known components: 0.96 ÷ (0.992 × 0.991 × 0.994 × 0.992) = 0.96 ÷ 0.9694 ≈ 0.990, making C correct.

Why the distractors fail:

• B (0.969) is the most tempting trap - it's approximately the product of the four known components (0.9694), which is what you divide by, not the answer itself.
• D (0.992) is simply the reliability value assigned to two of the other components - there's no calculation justifying reusing it for R5.
• A (0.970) is close to the overall system reliability of 0.96, suggesting confusion between system-level and component-level values.

Memory tip: Think of it as "what's left over." Divide the target system reliability by what the other components collectively deliver - the quotient is the apportionment the remaining subsystem must achieve.

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