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Math Help - Quick prob question

  1. #1
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    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?
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  2. #2
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    Quote Originally Posted by aman_cc View Post
    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).
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  3. #3
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    Quote Originally Posted by Focus View Post
    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
    <br />
\mu(\cup_{n\geq 1} A_n)=\sum_{n\geq 1} \mu(A_n)<br />

    If it is a standard representation, do you mind pointing me where I can refer to these?
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  4. #4
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    Quote Originally Posted by aman_cc View Post
    Hi - I am sorry I do not understand the terms indicated in
    <br />
\mu(\cup_{n\geq 1} A_n)=\sum_{n\geq 1} \mu(A_n)<br />

    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
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  5. #5
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    Quote Originally Posted by Focus View Post
    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.
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