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You have a system with 6 components. In order for the system to work the following must be met:

Component 1 must work.
At least one of components 2, 3, 4 must work.
At least one of components 5, 6 must work.

Component 1 has an exponentially distributed mean lifetime of 1/2 a year.
Components 2,3,4 have an exponentially distributed mean lifetime of 1 year.
Components 5,6 have an exponentially distributed mean lifetime of 3/2 years.

All components function independently.

What is the probability that the system will function for at least 2 years?

So this is my thinking:

For each group of components, I find the probability that they will last less than 2 years using the cdf of exponential distribution. Then I subtract that probability from 1 to get the probability that they will last at least 2 years. If we define $p_i$ as the probability that the $i^{th}$ component will last 2 years, we have: \begin{align*} p_1 = \exp(-2\times 2) &= 0.018\\ p_2 = p_3 = p_4 = \exp(-1 \times 2) &= 0.135\\ p_5 = p_6 = \exp\{-2/3 \times 2\} &= 0.263. \end{align*} Then, using binomial probability, I find the probability that all of the components will break. Then, I subtract that from 1 to find the probability that at least one of the components will not break. \begin{align*} 1 - b(0;3,0.135) &= 1-(1-0.135)^3 &= 0.354\\ 1 - b(0;2,0.263) &= 1-(1-0.263)^2 &= 0.458 \end{align*} Finally, I multiply the 3 probabilities for the 3 different groups together to find my final answer. $$ (0.018)\times(0.354)\times(0.458) = 0.00296 $$ So am I right?

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    $\begingroup$ I had written it as a percent, but thank you for verifying. $\endgroup$ – picokol Sep 23 '15 at 15:57
  • $\begingroup$ OK, my idiosyncrasy: I dislike percents in probability statements... $\endgroup$ – Xi'an Sep 23 '15 at 17:55
  • $\begingroup$ Is this a homework question? If so, please add the self-study tag. $\endgroup$ – jlimahaverford Sep 28 '15 at 16:27
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100% correct, assuming that all of the components lifetimes are independent. I edited and added a little notation, partially so I can use it here. If I may give a couple suggestions.

  • Use more notation when you post here, because then people don't have to "back into" what value you are trying to calculate. Especially when you (correctly) are frequently flipping between a value and its complement.

  • In this case there was no need to use a binomial distribution. If the components within any of the groups had different survival probabilities, then you couldn't use the binomoal. Whereas, if you just said, the probability that the second group of components all fail, by independence is:

$$ (1-p_2) \cdot (1-p_3) \cdot (1-p_4). $$

  • Make sure you're aware of the assumptions you're making. You used independence, many times, and if you were not aware of that, it will come back to bite you.
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  • $\begingroup$ "All components function independently." & yes, I know of the alternate method to compute this. In fact, my reason for posting this was that I was getting different answers on both methods. I later realized that I had input 1 - b(0; 3, 0.263) instead of 1 - b(0, 2, 0.263) in Cran R... I just realized nobody caught on to that here. I'll edit that now as well. Edit: nevermind, I just noticed that you saw that. Thanks, I'm still new to this website $\endgroup$ – picokol Sep 28 '15 at 16:54
  • $\begingroup$ I had edited that, but it was awaiting peer review. $\endgroup$ – jlimahaverford Sep 28 '15 at 16:55
  • $\begingroup$ @picokol And thank you for pointing out the independence line. I was looking for it before but couldn't find it. It is important to know that using the binomial you are implicitly invoking that assumption. $\endgroup$ – jlimahaverford Sep 28 '15 at 16:58

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