1
$\begingroup$

If we define a set $[A,B]$ and a set $[a,b]$ which is contained inside the first set, what are the odds when picking a number from set $[A,B]$ that it is also contained inside the set $[a,b]$ if all numbers are equally likely to be picked?

If elements in our sets are $\in\mathbb N$, then we can simply divide the number of all elements we can pick from and the number of elements in our target set, making our odds:

$$p({\mathbb N})= \frac{b-a+1}{B-A+1}$$

For $\in\mathbb Z$, we simply watch out for the negatives:

$$p({\mathbb Z})= \frac{|b-a|+1}{|B-A|+1}$$

But what about when our sets are rational, $\in\mathbb Q$?
If we define our sets like here, we have:

$$p({\mathbb Q})=\frac{|b-a|+1+(|b-a|)\times n}{|B-A|+1+(|B-A|)\times n}, n\to\infty$$

If we take the limit as $n$ approaches $\infty$, we finally get:

$$p({\mathbb Q})=\frac{|b-a|}{|B-A|}$$

Is this a valid solution?

Also, I'm wondering how could one define these sets for $\in\mathbb R$ and calculate $p(\mathbb R)$?
What about $p(\mathbb C)$?

  • 0
    I don't understand the notation. Your set $(A,B)$ has two elements? Or do you mean something like $x\in \mathbb R$ such that $A≤x≤B$? – 2017-01-10
  • 0
    @lulu The second thing, edited the question replacing () with []. – 2017-01-10
  • 0
    But why not spell out exactly what you mean? I just guessed that you meant a subset of the reals....that's the sort of thing that should be stated. – 2017-01-10
  • 1
    To your question: I think you are asking about measure theory. If your sets have a nice geometry to them you can just take the ratio of the areas. If not, well you need something like a measure. There are some pretty bad sets out there...it's hard to come up with a notion that makes sense in all cases. – 2017-01-10

0 Answers 0