# Given sum of uniform random variables $Z_1 + Z_2 + \dots + Z_n =1$,what's the probability that $k$ R.Vs are at least $1/n$?

Given sum of uniform random variables on $[0,1]$, $Z_1 + Z_2 + \dots + Z_n = 1$, what is the probability that exactly k random variables are at least $\frac{1}{n}$?

In other words, what's

$Pr[\text{exactly k random variables are at least }\frac{1}{n} \mid Z_1 + Z_2 + \ldots + Z_n = 1]$

where $Z_i$s are uniform random variables on $[0,1]$?

Thanks!

• How are $Z_i$ sampled in a way that they are uniform random variables and they satisfy that sum criteria? The sum criteria makes them non uniform. – Hugh May 30 '16 at 10:45
• @Hugh I think it is conditional probability. – Element118 May 30 '16 at 10:49
• The posts means to say something like $P(\text{at least$kZ_i$are at least$1/n$}\mid Z_1+\dotsb+Z_n = 1)$ – Em. May 30 '16 at 10:50

The probability for $j$ particular variables to be at least $\frac1n$ is
$$\left(1-\frac jn\right)^{n-1}\;,$$
so by inclusion-exclusion the probability for exactly $k$ variables to be at least $\frac1n$ is
$$\sum_{j=k}^n(-1)^{j-k}\binom nj\binom jk\left(1-\frac jn\right)^{n-1}\;.$$