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The life span of a particular mechanical part is a random variable described by the following PDF: enter image description here

If three such parts are put into service independently at t=0, determine a simle expression for the expected value of the time until the majority of the parts will have failed.

I can get the PDF: $$ f_L(l) = 0.4 (0 \leq l \leq 2) \\ f_L(l) = -0.4l + 1.2 (2 < l \leq 3) $$ and the expectation: $$ E(l) = \int_0^3 l f_L(l) dl \approx 1.27 $$

I think 'majority' means 2 or more, so we can focus on two parts of the three, and pay no attention to the third. The translation is $E(max(l1, l2))$, how will this be derived I currently have no idea.


Sorry about the misleading remark "$E(max(l_1, l_2))$", it's wrong to neglect the third part, because if that one fails early, then we only need one of the rest to fail.

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    What are you thoughts?2012-12-16
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    Your question asks for the first time when 2 out of 3 components fail. You accepted an answer which studies the first time when 2 out of 2 components fail. This is quite different.2012-12-20
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    I agree with did. The accepted answer computes the first time when 2 out of 2 components fail.2012-12-21
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    CravingSpirit: Sorry but I am afraid that you lost me: you accepted an answer THEN offered a bounty, without unaccepting the answer nor commenting on the other answer? Could you keep me posted on the status of this question and the answers you received?2012-12-21
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    @did, sorry, I've been busy with other things. I will check both answers and give the bounty to the preferred one. Thanks!2012-12-24

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