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Question
A k-out-of-n system is one that will function if and only if at least $k$ out of the $n$ individual components in the system function. Assume that individual components function independently of each other. Assume also each individual component functions with probability $0.9$

Determine the long-run proportion of 3-out-of-5 systems that will function.

My Working
The binomial distribution looks to be a good fit here.
Summarising the data:

$p$ (the probability of success) $= 0.9$
$n$ (the total number of trials) $= 5$
$x$ (the number of successes needed) $= 3$

3-out-of-5 means the system needs at least $3$ components to function which means the system will function at $3$ components, $4$ components $5$ components.

Which in turn means the answer will be $1 - prob$($0$ components + $1$ components + $2$ components)

Therefore $1 - 0.0815$

= $0.9185$

Is this correct?

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    Thanks guys, indeed I made a calculation mistake. I did 1-binocdf(3,5,0.9) instead of 1-binocdf(2,5,0.9) - in matlab2011-10-26

1 Answers 1

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I do not at this point see what went wrong in the computation. The answer is $\binom{5}{3}(0.9)^3(0.1)^2+\binom{5}{4}(0.9)^4(0.1)^1+\binom{5}{5}(0.9)^5(0.1)^0.$ The calculator gives me an answer of $0.99144$.

Because it is easy to make a mistake, I also calculated using your (correct) idea. The answer then would be

$1-\left(\binom{5}{0}(0.9)^0(0.1)^5+\binom{5}{1}(0.9)^1(0.1)^4+\binom{5}{2}(0.9)^2(0.1)^3\right).$

Again, I get $0.99144$. This time, to make sure, I also did the calculation by hand.

Added: Maybe I have a good guess about what went wrong. Note that $\binom{5}{3}(0.9)^3(0.1)^2=.0729$. If we subtract this from $0.99144$, we get $0.91854$. So maybe in your calculation you used one term too many.

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    Thank you, I have inserted the parentheses.2011-10-26