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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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The reasoning is correct, but the calculation is not: the probability is $0.99144$. If you show the details of your calculations, we can see where you went astray. –  Brian M. Scott Oct 26 '11 at 5:40
    
You on the right track, but your numbers do not look right. $\mathbb{P}(X=0) = (1-p)^5 = 10^{-5}$, $\mathbb{P}(X=1) =5 (1-p)^4 p = 45 \cdot 10^{-5}$, $\mathbb{P}(X=2) = 10 (1-p)^3 p^2 = 810 \cdot 10^{-5}$. I am getting 0.99144 for the answer, instead of 0.9185. –  Sasha Oct 26 '11 at 5:41
    
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 matlab –  Arvin Oct 26 '11 at 6:05

1 Answer 1

up vote 3 down vote accepted

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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Thanks Andre, I made a silly error by doing 1-binocdf(3,5,0.9) instead of 1-binocdf(2,5,0.9) - in matlab –  Arvin Oct 26 '11 at 6:06
    
I had added that guess a few seconds ago, since I realized that that term probably accounted for the discrepancy. –  André Nicolas Oct 26 '11 at 6:08
    
What appears above is wrong. It says $1-\binom{5}{0}(0.9)^0(0.1)^5+\binom{5}{1}(0.9)^1(0.1)^4+\binom{5}{2}(0.9)^2(0.1‌​)^3$. It should say $1-\left(\binom{5}{0}(0.9)^0(0.1)^5+\binom{5}{1}(0.9)^1(0.1)^4+\binom{5}{2}(0.9)‌​^2(0.1)^3\right)$. –  Michael Hardy Oct 26 '11 at 10:14
    
Thank you, I have inserted the parentheses. –  André Nicolas Oct 26 '11 at 13:38

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