I've been trying to figure out a counting problem and can't wrap my head around how to calculate the probability.

If we let $X_{1}, . . . , X_{10}$ be independent random variable with a uniform(0,1) distribution. Find the probability that each of the intervals $[0.2i, 0.2i + .2)$, $i = 0, 1, 2, 3, 4$, contains exactly two of the Xi’s.

I know if we only had two random variables, the probability of landing in a specific interval would be $0.2 \times 0.2 = 0.04$ and in a random interval $0.04 \times 5$. But when we have ten random variables, I'm not sure how to proceed. I feel like maybe I need to incorporate some binomial coefficients perhaps?


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    $\begingroup$ You can look at this more easily as a multinomial distribution problem. en.m.wikipedia.org/wiki/Multinomial_distribution $\endgroup$ – Paul Feb 9 '16 at 0:32
  • $\begingroup$ That does make my life a lot easier! Would it then be: $P=(A=2, B=2, C=2, D=2, E=2) = \frac{10!}{5\times2!} \times 0.2^{10}$ Where A,B,C,D,E are the 5 intervals. $\endgroup$ – Tim Feb 9 '16 at 0:42
  • $\begingroup$ Close: $\;\mathsf P(A{=}2, B{=}2, C{=}2, D{=}2, E{=}2) \,=\, \dfrac{10!}{\color{crimson}{2!^5}} 0.2^{10}$. $\endgroup$ – Graham Kemp Feb 9 '16 at 2:25
  • $\begingroup$ Right! Thanks to both of you for your help. $\endgroup$ – Tim Feb 9 '16 at 14:34

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