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I have $n$ Bernoulli trials with probabilities of success and failure $p$ and $q=1-p$. How can I find a probability that number of successes is divisible by some given number. I.e. if we choose $3$ than we should consider cases where number of successes equals $0$, $3$, $6$, $9$, etc. Using standard formula for probability of exactly $k$ successes we get

$$ \binom{n}{0}p^0q^{n-0} + \binom{n}{3}p^3q^{n-3} + \binom{n}{6}p^6q^{n-6} + \dots = \sum\limits_k \binom{n}{3k}p^{3k}q^{n-3k} $$

Is there any way to do this summation and get some analytical expression? For other numbers, I believe, we should just change $k$ multiplier.

Boris
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  • Possible duplicate http://math.stackexchange.com/questions/918/how-do-i-count-the-subsets-of-a-set-whose-number-of-elements-is-divisible-by-3 – shyamupa Nov 29 '13 at 18:01

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Let $\omega=\exp\left(\frac{2\pi\mathrm i}m\right)$. Then

\begin{align} \sum_{k=0}^n\binom n{mk}q^{mk} &=\frac1m\sum_{j=0}^n\binom njq^j\sum_{l=0}^{m-1}\omega^{jl}\\ &=\frac1m\sum_{l=0}^{m-1}\sum_{j=0}^n\binom njq^j\omega^{jl}\\ &=\frac1m\sum_{l=0}^{m-1}\left(1+q\omega^l\right)^n\;. \end{align}

Thus

$$ \sum_k\binom n{mk}p^{mk}q^{n-mk}=\frac1m\sum_{l=0}^{m-1}\left(q+p\exp\left(2\pi\mathrm i\frac lm\right)\right)^n\;. $$

For $m=3$, this is

$$ \frac13\left(1+2\Re\left(\left(q+p\exp\left(\frac{2\pi\mathrm i}3\right)\right)^n\right)\right)\;. $$

For $m=4$, it's

$$ \frac14\left(1+(q-p)^n+2\Re\left(\left(q+\mathrm ip\right)^n\right)\right)\;. $$

joriki
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