# Probability theory. 100 items and 3 controllers

100 items are checked by 3 controllers. What is probability that each of them will check more than 25 items? Here is full quotation of problem from workbook: "Set of 100 articles randomly allocated to test between the three controllers. Find the probability that each controller has got to test at least 25 articles."

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This feels like a multinomial distribution. Each article has a 1/3 chance to be assigned to each controller. –  Ronald Apr 13 '12 at 15:26
nope, it's too easy –  Yurgen Apr 15 '12 at 17:45
anyway, the chosen answer matches with my solution. –  Ronald Apr 15 '12 at 19:25
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## 2 Answers

My try: $$\mathbb{P}(N_1\geq 25, N_2\geq 25, N_3\geq 25)=\frac{1}{3^{100}}\sum_{N_1\geq 25, N_2\geq 25, N_3\geq 25|N_1+N_2+N_3=100}\binom {100}{N_1,N_2,N_3}=$$ $$\frac{100!}{3^{100}}\sum_{N_1}\sum_{N_2}\sum_{N_3}\frac{1}{N_1!N_2!N_3!}=\frac{100!}{3^{100}}\sum_{N_1}\sum_{N_2}\frac{1}{N_1!N_2!(100-N_1-N_2)!}=$$ $$\frac{100!}{3^{100}}\sum_{N_1=25}^{50}\sum_{N_2=25}^{50-(N_1-25)}\frac{1}{N_1!N_2!(100-N_1-N_2)!}$$ You also may be interested in this and this

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Let $N_i$ denote the number of items checked by controller $i$. One asks for $1-p$ where $p$ is the probability that some controller got less than $k=25$ items. Since $(N_1,N_2,N_3)$ is exchangeable and since at most two controllers can get less than $k$ items, $p=3u-3v$ where $u=\mathrm P(N_1\lt k)$ and $v=\mathrm P(N_1\lt k,N_2\lt k)$.

Furthermore, $v\leqslant uw$ with $w=\mathrm P(M\lt k)$ where $M$ is binomial $(m,\frac12)$ with $m=75$ hence $w\ll1$. And $N_1$ is binomial $(n,\frac13)$.

Numerically, $u\approx2.805\%$ and $w\approx0.1\%$ hence $1-p\approx1-3u\approx91.6\%$.

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Could you explain me a little more about this answer? –  Pedro Tamaroff Apr 13 '12 at 17:26
@PeterT.off I did. What part remains mysterious to you? –  Did Apr 13 '12 at 17:27
@Dider Oh, I didn't notice the add. I'm still working on it. Probabilistic arguments are new to me. –  Pedro Tamaroff Apr 13 '12 at 17:30
thank you, it looks right –  Yurgen Apr 15 '12 at 17:39
Yurgen: The answer you were after is really the summation of 925 terms, with factorials and everything? (Which, from my answer, sums roughly to 0.9, but how to see this on the summation?) –  Did Apr 15 '12 at 18:18
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