0
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
1
$\begingroup$

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 $0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.