# what is the probability of choosing $m$ numbers from the interval $[0,1]$

As far as I know there are infinitely many real numbers between $[0,1]\subseteq R$, what is the probability of choosing a given set of numbers $\{x_1,...,x_m\}$ where $x_i\in[0,1]$ from $[0,1]$? where there is a uniform distribution over $[0,1]$.

-
$\displaystyle{\large 1}$. – Felix Marin Oct 1 '13 at 4:27
like the probability of choosing 5 real numbers from [0,1] is 1? – seteropere Oct 1 '13 at 4:28
Which is it, "$m$ numbers" (as in the title) or "a fixed number $m$"? – Robert Israel Oct 1 '13 at 4:31
@RobertIsrael Sorry, I mean choosing $x_1,...,x_m$ number from [0,1] – seteropere Oct 1 '13 at 4:34
The question was slightly changed. Now is $\displaystyle{\large 0}$. Just before "$\,\displaystyle{\large set}\,$" write "$\,\displaystyle{\large given}\,$". That will be clearer. – Felix Marin Oct 1 '13 at 6:39

If $x$ is a random variable selected according to a uniform distribution on $[0,1]$, then the probability that $x$ lies in the interval $I_{m,\epsilon}=[m-\epsilon, m+\epsilon]$ is exactly $2\epsilon$ (here, I assume that $\epsilon < m < 1-\epsilon$). As $\epsilon$ tends to $0$, the probability of $x$ lying within $\epsilon$ of $m$ tends to $0$ as well. In symbols, $$\lim_{\epsilon \to 0} P(x \in I_{m,\epsilon}) =\lim_{\epsilon \to 0} 2\epsilon =0,$$ which implies that the probability that $x \in I_{m,0}$ (i.e. $x=m$) is exactly $0$.

Here's another way to see this: suppose that $x$ was chosen according to a uniform distribution on $[0,1]$, and the probability that $x=m$ was $\epsilon >0$. Then $x \in [m,m]$ with probability $\epsilon$, so that $x \in [m+\delta,m+\delta]$ with probability $\epsilon$ as well (translation invariance). Over all possible $\delta$, we quickly add up to a probability greater than $1$.

Here's a third way to see this: if $x$ is taken as a random variable according to uniform distribution on $[0,1]$, then $x$ lies in an interval $[a,b]$ with probability $b-a$. That $x =m$ is equivalent to $x \in [a,b]$ with $a=b=m$, which gives a probability of $m-m=0$.

-
Nice..Thanks for the illustration. – seteropere Oct 1 '13 at 5:20

If you're using a uniform distribution on $[0,1]$, all numbers that you choose will be in $[0,1]$. But how many of those numbers you take is up to you. The distribution doesn't tell you how many numbers to take, it just tells you, e.g., that when you do choose a number $X$, the probability that $X < c$ is $c$ (where $c$ is any given number in $[0,1]$).

-
thanks.. so the probability of choosing $\{x_1,...,x_m\}$ numbers from [0,1] is 0. – seteropere Oct 1 '13 at 5:25