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In my perception, using the common sense, is less common, or less probable, to a random line be parallel that not to be, because to be parallel a line needs obey a restrictive rule. But anyone can, using simple probability, obtain that result:

specific line (r) = any line in R2 (e.g. y=0) amount of random lines (u) = infinite (inf) amount of parallel lines to r (s) = infinite (inf) probability of s parallel to r = s / u = inf/inf = undefined 

But there is another solution? Maybe using geometric probability, integrals or even empirical results?

[EDIT]

OK. I've understood the zero result and the "not impossible" thing. Thanks for answers and references.

But...

  • Is my initial perception wrong?
  • Is not easier to find a not parallel line instead a parallel?
  • Could someone prove or negate it?
  • If my perception is not wrong someone can calculate how easier is?
  • or maybe I really didnt understand the answers?

[NEW EDIT] So, is it the final answer? - Is my initial perception wrong? - A: No, its correct.

  • Is not easier to find a not parallel line instead a parallel?
  • A: Yes, it is easier.

  • Could someone prove or negate it?

  • A: Yes: '''The probability of finding a parallel line is zero, so the probability of finding a non-parallel line is equal to one. Since 1>0, you have the answer to your question. – Rod Carvalho'''

  • If my perception is not wrong someone can calculate how easier is?

  • A: No, nobody can because it is undefined. (?)

[FINAL EDITION]

Now I realize that the question resumes to: which is the probability of a random real number be equal to a specific other. Thanks for help.

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    The probability of finding a parallel line is zero, so the probability of finding a non-parallel line is equal to one. Since 1 > 0, you have the answer to your question.2012-09-14

3 Answers 3

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Suppose that we have a line $\mathcal{L}_1$ defined by

$\mathcal{L}_1 := \{ (x,y) \in \mathbb{R}^2 \mid y = a x + b \}$

and another line, $\mathcal{L}_2$, defined by

$\mathcal{L}_2 := \{ (x,y) \in \mathbb{R}^2 \mid y = \tan (\theta) x + c \}$

where $\theta$ is an observation of a random variable $\Theta$ uniformly distributed over $[0, \pi]$. Lines $\mathcal{L}_1$ and $\mathcal{L}_2$ will be parallel if $\theta = \tan^{-1} (a)$. However, since $\Theta$ has a continuous distribution, we have that

$\mathbb{P} \left( \Theta = \tan^{-1} (a) \right) = 0$

In other words, the probability that a line whose slope is randomly chosen is parallel to a given line is exactly equal to zero.

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    This clear now, thanks.2012-09-14
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A line in $\Bbb R^2$ is parallel to another line if and only if their slopes are equal, regardless of their intercept.

The probability that the slope $m$ of a random line is equal to the slope $m_f$ of some fixed line is therefore $P(m = m_f) = 0$, since $m_f$ is continuously distributed.

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    A probability of zero is not the same thing as an impossible event. For example, take a uniform random variable $u$ continuously distributed in $(0,1)$. What is the probability that $u = .56721$? The answer is $P(u=.56721) = 0$. In fact, the probability of $u$ assuming *any* value in the interval is zero. However, any sampling of $u$ must come up with a value in the interval, so clearly probability zero does not imply impossibility.2012-09-14
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The answer depends on the random distribution accordingto which the lines are produced. But if the probability of a random line being parallel to a given line $L$ is assumed not to depend on $L$ (a resonable symmetry assumption), then the assumption of a positive probability $p>0$ leads to a contradiction: If $p>0$ then there exists a natural number $n$ such that $n>\frac1p$. It is easy to construct $n$ differnt lines $L_1,\ldots, L_n$ through a single point $A$. The probability of a random line $L$ to be parallel to $L_i$ is assumed to be $p$ for all $i$. Since the events $L||L_i$ and $L||L_j$ are mutually exclusive, we obtain that the probability that $L$ is parallel to any of the $L_i$ is $p+p+\ldots+p$ with $n$ summands. This produces a total probability of $np>1$ - contradiction. We conclude that the assumption $p>0$ is wrong, hence $p=0$.