# probability of picking $\frac{1}{3}$ between the interval $[0,1]$?

My professor told us a situation in class which i do not understand very well.

He said "what is the probability of picking $\frac{1}{3}$ between the interval $[0,1]$?" and then he proved it to be $0$. He said something like if you continue to pick subsets of $\frac{1}{3}$ it will eventually branch off to $0$.

I wanted to ask my professor to clarify but he had a meeting afterwards. Therefore, I come to you all. Does anyone think they can explain this concept to me?

• There are infinite possible numbers (between 0 and 1) which can be picked. Thus the probability of picking a specific number is zero. – callculus Apr 12 '16 at 5:28
• If your professor means for the probability distribution over $[0,1]$ to be uniform, then he is correct that the probability of getting exactly $\frac{1}{3}$ is zero. One way to explain why this is the case is that measure of the set $\{ \frac{1}{3}\}$ is zero in this case. – Justin Benfield Apr 12 '16 at 5:28
• Roughly speaking, there are infinitely many points in $[0,1]$ and $\frac 13$ is just one of them, so the probability of picking it is $\frac 1{\infty}$. – BigbearZzz Apr 12 '16 at 5:28
• @JustinBenfield, i think my professor was using your idea – yaniz rakiov Apr 12 '16 at 5:30

$1/3$ lies in the interval $[0,1/2]$, and there is probability $1/2$ of any point we choose in $[0,1]$ lying in that interval. Further, $1/3$ lies in $[1/4,1/2]$, and there is probability $1/4$ of any point we choose in $[0,1]$ lying in this interval. Likewise, $1/3$ lies in $[1/4,3/8]$, which we would land in with probability $1/8$. Continue in this manner inductively constructing a sequence of nested intervals of length $1/2^n$, each containing the point $1/3$.
This means that the probability of choosing $1/3$ out of the interval $[0,1]$ is smaller than the length of any of these nested intervals that contain the point $1/3$, and hence the probability is smaller than $1/2^n$ for any positive integer $n$, so the probability must be $0$.
I will answer this question by asking you How many number are there in the interval $$[0,1]$$ and remember it is a continuous realnumber line