A certain light switch fails with a failure rate of 10 -4 . If the switch is used 20 times per week, what is the probability of the switch failing by the end of 3-year period?

Question

Options

• A7.32 ?10 -5
• B2.68 ?10 -1
• C3.12 ?10 -1
• D7.32 ?10 -1

• A
  3% (1)
• B
  83% (25)
• C
  7% (2)
• D
  7% (2)

Explanation

Option B is correct because the total number of switch uses over 3 years is 3 years × 52 weeks/year × 20 uses/week = 3,120 uses. Using the reliability formula, the probability of not failing across all uses is (1 − 10⁻⁴)^3120 ≈ e^(−0.312) ≈ 0.7319, so the probability of failing at least once is 1 − 0.7319 ≈ 0.268 = 2.68 × 10⁻¹.

Why the distractors fail:

• A (7.32 × 10⁻⁵): This ignores cumulative uses - it's essentially just the per-use failure rate scaled slightly, not compounded over 3,120 uses.
• C (3.12 × 10⁻¹): This uses np = 3,120 × 10⁻⁴ = 0.312 directly as the probability, skipping the final step of computing 1 − e^(−0.312).
• D (7.32 × 10⁻¹): This is the survival probability e^(−0.312) ≈ 0.732 - the probability the switch doesn't fail - which is the complement of what the question asks.

Memory tip: Think "Survive then Flip" - first calculate the probability of surviving all uses with R = e^(−λn), then flip it (1 − R) to get the failure probability. If your answer looks like it matches one of the other choices, you likely stopped one step too early.

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