The length of life of electronic component is exponentially distributed with mean 100 hr. A system has n = 5 components, which operate independently. The system fails if at least two components fail. What is the probability that system will survive 100 hr?

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1 Answer 1


There are $5$ iid $RV_s$ : $X_1 X_2 X_3 X_4 X_5$

Let $T_1$ be the lifetime for the first failing component.

Let $T_2$ be the lifetime for the second failing component.

Noticing that $Pr(T_2 \ge s + t|T_1 \ge t) = Pr(T_2\ge s)$ (Memoryless)

$Pr(T_1 \ge t) = Pr(min\{X_1, X_2, X_3, X_4, X_5\}\ge t) = exp({-5\lambda t})$

Similarly, we have:

$Pr(T_2 \ge t) = exp({-4\lambda t})$

The probability desired is computed as:

By Law of Total Probability $\Rightarrow$

$Pr($Systems survives 100 hours $\land$ $\exists$ one component fails within 100 hours = True$)$ + $Pr($Systems survives 100 hours $\land$ $\exists$ one component fails within 100 hours = False$)$


$\int_{0}^{100} Pr(T_1 = t)Pr(T_2 > 100-t)dt + exp(-5)$

where $exp(-5)$ is computed from:

$Pr($ All five components survive 100 hours$) = \prod_{i=1}^{5} Pr(x_i\ge 100) = exp(-1)^5$


$(-5exp(-4))\int_{0}^{100} exp(\frac{-t}{100})dt + exp(-5)\approx 0.0646264064$

                                              Hoping it will help. 
                                                   KeCen Zhou
  • $\begingroup$ The answer my Professor gave was 0.0646. Any idea how that was obtained? $\endgroup$
    – Jebediah
    Feb 28, 2014 at 3:26
  • $\begingroup$ @Jebediah No idea ... Can you put your prof's answer? I'm curious to see how. $\endgroup$
    – MFSO1991
    Feb 28, 2014 at 3:41
  • $\begingroup$ Attached image above is the solution. $\endgroup$
    – Jebediah
    Feb 28, 2014 at 3:51

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