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Part CCSIR NET June 2025parallel-is-the-maximum-so-the-failure-probabilities-multiply-not-the-survivals

Parallel is the maximum so the failure probabilities multiply not the survivals

A system has two components C1C_{1} and C2C_{2} put in parallel. The components C1C_{1} and C2C_{2} have independent lifetimes X1X_{1} and X2X_{2}, respectively. The probability distribution of X_j is exponential with mean 1/j, j = 1, 2. Suppose that R(t) and h(t) are the reliability and the hazard rate functions of the system, respectively. Then, which of the following statements are true?

  1. A.R(t) = e−2te^{-2t} + e−te^{-t} − e−3te^{-3t}, t > 0✓ Correct
  2. B.The expected lifetime of the given system is 1
  3. C.h(1)=(2e+e2−3)/(e+e2−1)h(1) = (2e + e^{2} - 3)/(e + e^{2} - 1)✓ Correct
  4. D.h(3)=(4e3+e6−5)/(e3+e6−1)h(3) = (4e^{3} + e^{6} - 5)/(e^{3} + e^{6} - 1)

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests execution slip.

See pricing

50 are analysed free — try those first.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Related counterexample: The maximum of independent exponentials is exponential

More on this topic

The chapter behind this: The standard distributions — free to read

From Probability › Standard discrete and continuous distributions

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

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