I've been asked the following question:

if you pick any rational number r in the interval from [0, 1] that the probability of picking this rational number is 0.

I've seen three different ways to solve this problem:

One involving how the probability of picking an irrational number is 1 and the probability of picking a rational number is 0.

Another where the length from a smaller interval contained within [0,1] and also contains r is 2m but since the length from [0,1] is 1. then the probability of picking a rational number is 2m/1=2m. However, i dont get the rest of this way. The rest of this solution involves picking even smaller intervals and the probability of picking a rational number is found by 2m where m approaches 0. i'm not understanding how you can get the probability to be 0 using this method.

Lastly, there are infinite rational numbers between 0 and 1 and since you want to pick 1 number that is the same as saying the probability of picking a rational number is the same as 1/x where x approaches infinity.

Can anyone clarify these three methods? Especially the 2nd one!

  • $\begingroup$ If we specify that all rationals in our interval are equally likely, then yes the probability is $0$. However, if they are not all equally likely, we could even have all rationals have non-zero probability. $\endgroup$ – André Nicolas May 12 '16 at 1:35
  • $\begingroup$ they are equally likely $\endgroup$ – kero May 12 '16 at 1:38
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    $\begingroup$ If they all have probability $a\gt 0$, find an $N$ such that $Na\gt 1$. Then if $S$ is any set of rationals with $N$ elements, then $\Pr(S)= Na\gt 1$, impossible. $\endgroup$ – André Nicolas May 12 '16 at 1:44

You didn't specify any distribution, so the probabilities could be whatever we want them to be. The question makes sense e.g. if you specify the uniform distribution over the interval $[0,1]\subset\mathbb R$.

For the second method: Choose some bijection $f:\mathbb N\to\mathbb Q$ between the natural numbers and the rational numbers and consider the set

$$ [0,1]\cap\bigcup_{n\in\mathbb N}[f(n)-m2^{-n},f(n)+m2^{-n}]\;. $$

This set contains all rational numbers in $[0,1]$, its measure is at most $2m\sum_{n\in\mathbb N}2^{-n}=2m$, and we can choose $m$ arbitrarily small. It follows that the probability to choose a rational number is $0$.


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