Independent Events

Question

Three independent switches in a circuit close successfully with probabilities 0.9, 0.8, and 0.7 respectively; what is the probability that at least one of these three switches fails to close?

RankGuru · Accepted Answer

For independent events the probability that all occur equals the product of their individual probabilities, and the complement rule converts an "at least one fails" question into one minus "all succeed". This complement-with-independence pattern is a staple of JEE Advanced reliability problems. All three switches close successfully with probability 0.9 × 0.8 × 0.7 = 0.504. The event "at least one fails to close" is exactly the complement of "all three close", so its probability is 1 − 0.504 = 0.496. Option 0.504 mistakenly reports the all-succeed probability rather than its complement. Option 0.6 wrongly adds individual failure probabilities 0.1 + 0.2 + 0.7 with errors. Option 0.244 attempts to multiply failure probabilities 0.1 × 0.2 × 0.7 and confuses "all fail" with "at least one fails". The reasoning rests on the multiplication rule for independent events combined with the complement principle P(at least one) = 1 − P(none of that type). Plausibility check: 0.496 is a valid probability in (0,1), and it should be slightly below 0.5 because each switch is fairly reliable, which matches intuition that failure of some switch is roughly a coin-flip-level risk.

Independent Events

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