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Here is the actual question: $A$ is random variable representing the lifespan of a component. It is an exponential law with an average of 10. Considering a system with $n$ components $A$, what is the minimum value of $n$ so that the system has a fiability of 0.999 for a 4 year period?

I thought about approximating it with the normal law (even if it's probably a pretty bad approximation). This gave me this random variable $Z_n = (Y-\nu)/(\sigma \sqrt{n})$.

Then I figured I was looking for $\mathbb P(Y \geq 4) = 1 - \mathbb P(Y \leq 3)$. In this case $\mathbb P(Y \leq 3)$ would be 0.001 since $\mathbb P(Y \geq 4) = 0.999$. With the cdf table I find that it gives -3.09. So I solve the equation $(3-10n) / (10 \sqrt{n}) = -3.09$. This gives an answer of 10 (around that). The actual answer is 7.

What have I done wrong and how could I find the answer of that question?

Edit : The components are independent from each other; if only one works, the system works. So basically only one of the components needs to have a life span of 4 years. The problem doesn't mention if the components are replaced so I assumed they weren't.

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    If the system fails before $x$ years, then that means that *all* of the components failed before $x$ years. (Right?) And, the latter is the *joint* probability that each component has a lifetime of less than $x$. Now, to continue, think about the extra structure you have in your problems. Namely, can you rewrite the right-hand side using assumptions about the $A_i$?2011-10-30

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If (and this is a big if) what you mean is that the $n$ components all start at time zero, that each component has a lifetime exponentially distributed with mean $10=1/\lambda$ and if the question is to know for which minimal value of $n$ one has $p_n\leqslant\varepsilon$ where \varepsilon=1/1'000 and $p_n$ is the probability that none of the $n$ components is still in function at time $t=4$, then here is your answer.

The probability that a given component is not functioning at time $t$ is $p_1=1-\exp(-\lambda t)$ hence the probability that none of them is functioning is $p_n=(p_1)^n=(1-\exp(-\lambda t))^n$. Numerically, $p_6=0.0013$ and $p_7=0.0004$ hence $p_6>\varepsilon>p_7$ and the minimal $n$ is $7$.

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    No it is not, see [here](http://en.wi$k$ipedia.org/wiki/Exponential_distribution).2011-10-30
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For a continuous random variable:

$P(Y\le 3) \neq P(Y<4) \; ,$

therefore use

$P(Y\ge 4) = 1 - P(Y < 4) \; .$

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    Yes, that is also incorrect, since your $Y$ variable as you constructed it is a kind of average lifetime, but what is needed is looking at the maximum lifetime. See Didier's answer.2011-10-30