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Suppose a number, $x$, is picked randomly from the interval $[0,1]$. What is the probability that $x=1/10$? What is the probability that $x=m/10$ for some $m=0,....,10$. What is the probability that $x=m/100$ for some $m=0,...,100$? What is the probability that $x=m/n$ for some $m \le n$? What is the probability that x is rational?

Not sure how to do this. I know with an increasing sequence, the probabilities of $P(A_n)$ grow with $n$ and approach the union and with a decreasing sequence the probabilities of $P(A_n)$ get smaller with $n$ and approach the intersection. How is this useful in the given problem?

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This doesn't make sense. The probability of $x$ being any particular number on the interval is $0$. –  Eric Angle Nov 19 '12 at 21:59
    
@Eric Angle-Why is the probability 0? If the number is on the interval, then shouldnt there be some chance it will occur? –  user50015 Nov 19 '12 at 22:59
    
Eric Angle is right; read my answer. Note the p.s. at the bottom. –  Josh Keneda Nov 19 '12 at 23:50
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A number can't be "picked randomly" unless we specify some probability distribution. Informally, the distribution tells us whether any values are more likely to be chosen than others. Read the first few sentences of this for more info.

We'll use the continuous uniform distribution on $[0,1]$. Normally, if people say that they've chosen a "random" number, this is the distribution that's implicitly being used. It means that all values have the same probability of being chosen. It also means that, if we take an interval $[a,b] \subset [0,1]$, the probability that we choose a number in $[a,b]$ is $b-a$, the length of the interval. So the probability that we pick a number in $[0, \frac{1}{2}]$ is $\frac{1}{2}$. I assume this is the distribution we're using; otherwise, we can work it out for another distribution. But understand that there's no such thing as a "random number" on $[0,1]$ without any probability distribution.

Now, let $x \in [0,1]$, and let $P(x)$ denote the chance that $x$ is chosen. I claim that the probability of choosing $x$ is $0$. There are two ways of seeing this. First, note that, if it were nonzero, then we have a problem. Since the distribution is uniform, all other points have the same probability of being chosen as $x$. So the probability that we pick a number in $[0,1]$ is the number of points in $[0,1]$ times $P(x)$, the probability of choosing $x$. But there are infinitely many points in $[0,1]$. So the probability of picking a number in $[0,1]$ is $\infty$. But probabilities should be between 0 and 1. So this is a problem. Each individual point must be chosen with probability 0.

Another way of seeing it is this: for any $\epsilon>0$, we can cover $x$ with an interval of length $\epsilon$. So $P(x)\le \epsilon$. But $\epsilon$ was arbitrary. So $P(x) = 0$.

Using this strategy, we can see that the probability that a rational number is chosen is $0$.

Let $\epsilon>0$. Since $\mathbb{Q}$, the set of rational numbers, is countable, we can enumerate it, i.e. $\mathbb{Q}=\{r_i\}_1^\infty$. Cover the $n$th rational with an interval of length $\frac{\epsilon}{2^n}$. Then the probability that we choose something in $\mathbb{Q}$ must be less than a sum of the lengths of the intervals that cover it. So, if $P(\mathbb{Q})$ denotes the probability that a rational is chosen, we have: $P(\mathbb{Q})\le \sum_1^\infty \frac{\epsilon}{2^n} = \epsilon$. So $P(\mathbb{Q})\le \epsilon$. But $\epsilon$ was arbitrary, so this can only be true if $P(\mathbb{Q})=0$. (This is what I meant on my last post when I said that $\mathbb{Q}$ has Lebesgue measure 0. Don't worry about that terminology, though. It's not important in this context.)

Let me know if that helps.

P.S. - It also might be important to note that $P(x)=0$ does not mean that $x$ cannot be chosen. If that confuses you, read about events that occur almost surely or this blog post.

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Just assuming a random number, not necessarily uniformly distributed and yes $m$ and $n$ are integers. –  user50015 Nov 19 '12 at 21:49
    
-I dont know what a Lebesgue measure is so could explain why that is relevant in calculating the probability of each case? –  user50015 Nov 19 '12 at 22:57
    
I'll edit my original response to see if I can clarify some points. :) –  Josh Keneda Nov 19 '12 at 23:20
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