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An electric system is made up of $k$ modules. In each module, there are $N$ resistors in series. What is the expected value of working modules if $m$ resistor break down? A module is not working if there is at least $1$ bad resistor in it. enter image description here

I can calculate the trivial cases (for example, $m=1$, or $m>(k-1)*N$), but I can't find out the general solution.

For $m\leq N$, I think I can use the binomial distribution: The probability for a module to have $n$ bad resistor is $$p(n)=\binom{N}{n}\left(\frac{1}{N}\right)^n\left(1-\frac{1}{N}\right)^{N-n}$$ and we would need to sum up this probability to cover all of the possible cases and multiple them with $1, 2, \dots N$ to get the expected value, but it seems to be a really big sum.

Bounty: I'd really like to have a solution to this problem, so I started a bounty.

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Botond
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1 Answers1

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The probability the first module has no bad resistors is the probability the $m$ bad resistors are in the other $k-1$ modules which by a counting argument is $\dfrac{{N(k-1) \choose m}}{{Nk \choose m}}$ - and you can take this to be $0$ if $m \gt N(k-1)$

So the expected number of working modules is $$k\dfrac{{N(k-1) \choose m}}{{Nk \choose m}}$$

Henry
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  • It might be trivial, but why is the expected value is the product of the number of the modules and the probability of 1 working module? – Botond May 22 '18 at 12:59
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    @Botond strictly speaking the expected number of first modules working is the probability it is working and similarly the expected number of second modules working and so on, so the same expression, and the final result is then the sum of $k$ copies of $\frac{{N(k-1) \choose m}}{{Nk \choose m}}$, one copy for each module, by linearity of expectation – Henry May 22 '18 at 14:12
  • Thank you, I think I start to understand it, but I will need a bit more time to understand it completely. – Botond May 22 '18 at 14:27