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There is a line of $10$ units length.I am marking a point on the line What is the probability that the point is exactly on the middle of the line ?

Can we apply classical probability here : $P(A)=\frac{n(A)}{n(S)}$ where $A$ is the event and $S$ is the sample space .Then it will give the answer $\frac{1}{\infty}$.

Instead of this approach I use another method.On the line I marked $10$ points including middle point.Selecting middle point from these $10$ points will be $0.1$. If we increase the points to $100$ the probability reduces to $0.01$. As the number of points increases probability tends to zero.

Now can we say that probability is zero ? If so, using language of probability bisecting line becomes an impossible event. Does the problem lies in classical approach ?

Madhu
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1 Answers1

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The probability of selecting any specific point, including the midpoint, is zero. That doesn't mean it's impossible, after all when we pick a point at random some point is selected, so probability zero events occur all the time.

As for the implication that bisecting the line is impossible, when we draw the bisector we are not selecting an intersection at random, but by design. So it is correct to say it is "impossible" to bisect a line segment on accident (at random), but it is surely possible to do it on purpose.

vadim123
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    Isn't it contradictory to say that probability-0 events happen all the time? – 355durch113 Oct 13 '15 at 04:15
  • Your answer is contradictory. Definition of probability says that if probability is zero, the event is impossible. But in the next line, you say that it isn't. – Aditya Agarwal Oct 13 '15 at 04:16
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    Actually, in sample spaces with uncountably many elements, mathematicians are forced to distinguish between "impossibility" and "having probability zero". Read the answer to this question: http://math.stackexchange.com/questions/155156/is-it-generally-accepted-that-if-you-throw-a-dart-at-a-number-line-you-will-neve – balddraz Oct 13 '15 at 04:21
  • @ZeroXLR. I don't know what measure theory is. But I will stick to my point. Because from the infinite set of the numbers in $[0,10]$, one number, that is $5$ will always be "out" in the numerator of $\frac{n(A)}{n(S)}$, not letting the probability ever be $0$. It is asymptotic. – Aditya Agarwal Oct 13 '15 at 04:29
  • @AdityaAgarwal My comment has nothing to do with your answer below. It is neither a response to nor an approval of it. I leave it to the others to decide on that matter. Instead, my comment is in response to your own comment above in this post where you define "impossibility" as "having probability zero". I merely pointed out that such a definition meets difficulties for very large sample spaces. – balddraz Oct 13 '15 at 04:35
  • @ZeroXLR, I also replied to your reply to my comment on this post. Don't misinterpret. – Aditya Agarwal Oct 13 '15 at 04:38
  • @ZeroXLR Why do u emphasise uncountable? – Vim Oct 13 '15 at 04:42