1. ## Quick prob question

I believe this might be more of a analysis question. (@admin - if you feel the same you may please move it accordingly. Thanks)

We pick a real number randomly from [0,1]. What is the probability it is rational?

I think it is 0. (because of the cardinality stuff) But how to demonstrate it formally?

2. Originally Posted by aman_cc
I believe this might be more of a analysis question. (@admin - if you feel the same you may please move it accordingly. Thanks)

We pick a real number randomly from [0,1]. What is the probability it is rational?

I think it is 0. (because of the cardinality stuff) But how to demonstrate it formally?
What is the probability that you pick an x that I fix? Use the additivity of the measure $\mu(\cup_{n\geq 1} A_n)=\sum_{n\geq 1} \mu(A_n)$ to calculate the probability of it being a rational (hint; rationals are countable).

3. Originally Posted by Focus
What is the probability that you pick an x that I fix? Use the additivity of the measure $\mu(\cup_{n\geq 1} A_n)=\sum_{n\geq 1} \mu(A_n)$ to calculate the probability of it being a rational (hint; rationals are countable).
Hi - I am sorry I do not understand the terms indicated in
$
\mu(\cup_{n\geq 1} A_n)=\sum_{n\geq 1} \mu(A_n)
$

If it is a standard representation, do you mind pointing me where I can refer to these?

4. Originally Posted by aman_cc
Hi - I am sorry I do not understand the terms indicated in
$
\mu(\cup_{n\geq 1} A_n)=\sum_{n\geq 1} \mu(A_n)
$

If it is a standard representation, do you mind pointing me where I can refer to these?
http://en.wikipedia.org/wiki/Measure_(mathematics)

The idea is that if $A_n$ are disjoint sets, then you can represent the probability of the countable union as the probability of the sum. So when you know the probability of picking a fixed $x\in [0,1]$, you can just do this
$\mathbb{P}(\mathbb{Q})=\mathbb{P}(\bigcup_{q \in \mathbb{Q}}\{q\})=\sum_{q\in \mathbb{Q}}\mathbb{P}(\{q\})$

I hope this is helpful in some way

5. Originally Posted by Focus
http://en.wikipedia.org/wiki/Measure_(mathematics)

The idea is that if $A_n$ are disjoint sets, then you can represent the probability of the countable union as the probability of the sum. So when you know the probability of picking a fixed $x\in [0,1]$, you can just do this
$\mathbb{P}(\mathbb{Q})=\mathbb{P}(\bigcup_{q \in \mathbb{Q}}\{q\})=\sum_{q\in \mathbb{Q}}\mathbb{P}(\{q\})$

I hope this is helpful in some way
Thanks Focus. Let me read it carefully and I'll post any further questions. I have not read all this before, so I guess will take sometime.