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Here are some questions I have and I can't ansewer

  1. Assume that we pick random a positive integer $n$. What is the probability

    (a) n=10

    (b) $n$ is even

    (c) $n$ is a prime number

  2. If we pick random a real number what is the probability this number is rational ?

Any help?

Thank's in advance!

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    What is a random integer? Do you assume any distribution?2012-04-18
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    @dtldarek: I mean we pick random a positive integer2012-04-18
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    How could $P(n = 10)$ possibly be greater than zero (with a uniform distribution)?2012-04-18
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    One cannot pick an integer at random with all integers equally likely, at least if as usual we want probabilities to be countably additive. So a distribution has to be specified. If you are interested in the densities (which are not probabilities) for $10$ we get $0$, for even we get $1/2$, for prime we get $0$.2012-04-18
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    @passenger: The problem with your question is that "we can't really pick a random positive integer". If we could and if we could talk about probabilities, then the probability for getting a specific number would be zero. But if you then sum up all the probabilities then you should get the probability of getting just any integer which should be one, but is zero.2012-04-18
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    @TheChaz: I am sure that this probability is zero but I can't prove it.2012-04-18
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    @passenger: So what people are saying is that the probability if not well defined. So in a sense, the question doesn't make sense.2012-04-18
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    Would the answers $2^{-10}$ and $\frac{2}{3}$ for 1a) and 1b) satisfy you?2012-04-18
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    @AndréNicolas: Can you explain a little more how did you found these probabilities ? And one more question: "One cannot pick an integer at random with all integers equally likely at least if as usual we want probabilities to be countably additive." why ?2012-04-18
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    Let's say that each integer has a probability $a > 0$ of being selected. The Archimedian property states that if we take enough of these probabilities and add them up, we'll get above 12012-04-18
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    @passenger: They are not probabilities, they are asymptotic densities. For $\{10\}$, if $n \ge 10$ we will have $|S_n|=1$ and therefore $|S_n|/n=1/n$. But $1/n\to 0$ as $n\to\infty$. For the evens, if $S_n$ is $n/2$ if $n$ is even, and $n/2-1$ if $n$ is odd. Divide by $n$. The limit is $1/2$. For the primes, I used the *Prime Number Theorem*, though that's overkill and one can do it with less machinery.2012-04-18
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    O.K Thank you all for your time!2012-04-18

2 Answers 2

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As has been pointed out in the comments, there is no probability distribution on the positive integers that assigns equal weight to every integer.

But let $S\subseteq \mathbb{N}$ be a set of positive integers, and for every positive integer $n$, let $S_n$ be the set of all $k\in S$ such that $k\le n$. Let $|S_n|$ be the number of elements in $S_n$. Then $$\lim_{n\to\infty} \frac{|S_n|}{n},$$ if it exists, can be viewed as a measure of how "large" $S$ is. By that criterion, the answer for $S=\{10\}$ is $0$. The answer for $S$ the even numbers is $1/2$, while the answer for the primes is $0$.

However, $\lim_{n\to\infty}\frac{|S_n|}{n}$ need not exist. Moreover, even if we restrict attention to subsets of $\mathbb{N}$ for which the limit exists, this limit is not a probability distribution.

On the reals, there is no probability distribution that gives equal weight to all intervals of (say) length $1$. But let's restrict attention to a specific interval, say $[a,b]$, and use the uniform distribution on this interval. Then the probability that a randomly chosen point in $[a,b]$ is rational turns out to be $0$. Almost all real numbers are irrational, indeed almost all real numbers are transcendental.

For a proof that the rationals form a negligibly small subset of, say, $[0,1]$, let $\epsilon>0$. The rationals in $[0,1]$ form a countable set, so they can be listed as $r_1,r_2,r_3,\dots$. Put an interval of width $\frac{\epsilon}{2^1}$ about $r_1$, an interval of width $\frac{\epsilon}{2^2}$ about $r_2$, and so on. The sum of the lengths of these intervals is $\epsilon$. So the rationals are a subset of sets of arbitrarily small measure.

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    $\lim_{n \to \infty} a_n$ is sometimes called the "asymptotic density", "natural density", or "arithmetic density" of $S$.2012-04-18
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    @AndréNicolas Yeah, I know that already (therefore I deleted the comment), but the part "let $S$ be a set of positive integers" can still confuse someone, "set of some positive integers" would be more clear to me, or at least please add $S \subseteq \mathbb{N}$ there.2012-04-18
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    @dtldarek: Thanks, I added the $S\subseteq \mathbb{N}$ that you suggest.2012-04-18
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The densities given by André Nicolas can be used as a measure of the required probabilities. For the first case the probability is zero as a limit (infinitismal probability) but when infinitely added it gives 1. This is just like the area of a strip with the width dx , the area of the strip is zero as a limit (infinitismal) but when infinitely added (the integration) it gives a value which is the area under the curve.