Assume a continuous distribution like a normal distribution. In that the probability of a point "x" is 1/infinity since we have infinite points. Therefore , for such cases we try to find probability of "x" to "x+dx". Also, if 1/infinity means impossible then probability of every point in normal will be impossible and hence their sum be impossible which is not true. If the distribution is continuous then we find the probability of a range like "x" to "x+dx" and if it is discrete, however large be the sample space, you will get a value however small it be.

The concept of probabilities is often not well understood, due to the big differences in the discrete case and the continuous case, which is best explained through calculus. There is a subtle difference in the Probability Distribution Function (pdf), which is similar to how Riemann sums are the discrete analogue of integration.

For example, it is not true that "normal distribution ... probability of a point \(x\) is 1/infinity SINCE we have infinite points". It is possible to have infinite points, and also a non-zero probability for a given point. For example, consider the action of flipping a fair coin. If it lands heads, we write 0. If it lands tails, we chose from a uniform distribution on \( [0,1]\). Then, with probability \( \frac {1}{2} \), the value is 0, dispute having infinitely many possible values due to the uniform distribution.

In order for the concept of probability = \( \frac {1}{N} \) to make sense, we are looking at a uniform distribution on \( N\) points. However, this does not extend to the case of \( N = \infty\). There is no such thing as the uniform distribution on the integers (discrete case) or real numbers (continuous case). The reason gets technical, so I'm give a simplistic idea.

Consider the discrete case, where \(X\) is our random variable of uniform distribution on integers. What is the probability that ( X=1)? If it is 0, then the total probability on \(X\) is 0 (since the countable sum of 0 is 0). If it is not 0, then the total probability on \(X\) is infinite (since the infinite sum of equal positive numbers is infinite). This contradicts the fact that the cumulative distribution frequency of a probability distribution must be 1.

Somewhat interestingly, while we can talk about the uniform random distribution on \([0,1]\), we cannot talk about the uniform random distribution on the reals.

For all practical purposes 1 / \( \infty \)= 0. Any number divided by \( \infty \) can be considered as a zero,because for any number \( \frac {a} {b} \), as \(b\) approaches \(\infty\) the number approaches zero. Imagine dividing a hypothetical continuous bread piece after piece,the pieces of bread you reach once you have divided it an infinite number times is so infinitesimal it can be considered to be a zero. Therefore yes practically speaking,the probability of an event occuring 1 / \( \infty \) times is zero(Which in probability terms is impossible or implies an infinite sample space).
EDIT: Strictly speaking infinity is not a number but rather an idea,so it would be senseless to perform any arithmetic operations on infinity because that would lead to specious conclusions and would be mathematically invalid. Thus \(0\) is merely the limit of the expression \( \frac {a} {b} \) as b approaches \(\infty\) .

I have some thoughts on this and wanted to share them. Feel free to be critic.

You have to be careful when trying to evaluate something like 1/infinite. First you have to be much more precise, like: lim 1/N where N approaches infinity. In that case, you would want to say it is 0 compared to some finite probability, like 1/2, but still that isn't that easy. Take this for example:
You go in front of your house and take a coin to toss it. If you get heads, you'll stop there. If you get tails, you will go on a 10km walk around the city. Let's define a walk to be unique if it has at least one tiny movement different than some other walks or there is a unique view that you see while walking, and there are obviously infinite many walks, each with probability 1/inf.

Now, there are 1/2 chance that you'll end up walking an "impossible" walk :D

Is 0.00000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000.......................00000000000000 0000000001 equal to 0?

If there are infinitely many zeros? Yes. Of course it is, for the same reason that .999...=1. Here's an intuitive argument:

Suppose that the number you have exists (and it does because completeness of the reals and it's lower bounded). Call it x. If there are infinitely many zeros, 10x also has inf. zeros before the one. So 10x=x implies x=0.

"If probability of some event occurring is lim 1/x as x -> infinity" is equivalent to saying "If probability of some event occurring is 0". Yes. The event is impossible.

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TopNewestAssume a continuous distribution like a normal distribution. In that the probability of a point "x" is 1/infinity since we have infinite points. Therefore , for such cases we try to find probability of "x" to "x+dx". Also, if 1/infinity means impossible then probability of every point in normal will be impossible and hence their sum be impossible which is not true. If the distribution is continuous then we find the probability of a range like "x" to "x+dx" and if it is discrete, however large be the sample space, you will get a value however small it be.

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The concept of probabilities is often not well understood, due to the big differences in the discrete case and the continuous case, which is best explained through calculus. There is a subtle difference in the Probability Distribution Function (pdf), which is similar to how Riemann sums are the discrete analogue of integration.

For example, it is not true that "normal distribution ... probability of a point \(x\) is 1/infinity SINCE we have infinite points". It is possible to have infinite points, and also a non-zero probability for a given point. For example, consider the action of flipping a fair coin. If it lands heads, we write 0. If it lands tails, we chose from a uniform distribution on \( [0,1]\). Then, with probability \( \frac {1}{2} \), the value is 0, dispute having infinitely many possible values due to the uniform distribution.

In order for the concept of probability = \( \frac {1}{N} \) to make sense, we are looking at a uniform distribution on \( N\) points. However, this does not extend to the case of \( N = \infty\). There is no such thing as the uniform distribution on the integers (discrete case) or real numbers (continuous case). The reason gets technical, so I'm give a simplistic idea.

Consider the discrete case, where \(X\) is our random variable of uniform distribution on integers. What is the probability that ( X=1)? If it is 0, then the total probability on \(X\) is 0 (since the countable sum of 0 is 0). If it is not 0, then the total probability on \(X\) is infinite (since the infinite sum of equal positive numbers is infinite). This contradicts the fact that the cumulative distribution frequency of a probability distribution must be 1.

Somewhat interestingly, while we can talk about the uniform random distribution on \([0,1]\), we cannot talk about the uniform random distribution on the reals.

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For all practical purposes 1 / \( \infty \)= 0. Any number divided by \( \infty \) can be considered as a zero,because for any number \( \frac {a} {b} \), as \(b\) approaches \(\infty\) the number approaches zero. Imagine dividing a hypothetical continuous bread piece after piece,the pieces of bread you reach once you have divided it an infinite number times is so infinitesimal it can be considered to be a zero. Therefore yes practically speaking,the probability of an event occuring 1 / \( \infty \) times is zero(Which in probability terms is impossible or implies an infinite sample space). EDIT: Strictly speaking infinity is not a number but rather an idea,so it would be senseless to perform any arithmetic operations on infinity because that would lead to specious conclusions and would be mathematically invalid. Thus \(0\) is merely the limit of the expression \( \frac {a} {b} \) as b approaches \(\infty\) .

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I have some thoughts on this and wanted to share them. Feel free to be critic.

You have to be careful when trying to evaluate something like 1/infinite. First you have to be much more precise, like: lim 1/N where N approaches infinity. In that case, you would want to say it is 0 compared to some finite probability, like 1/2, but still that isn't that easy. Take this for example: You go in front of your house and take a coin to toss it. If you get heads, you'll stop there. If you get tails, you will go on a 10km walk around the city. Let's define a walk to be unique if it has at least one tiny movement different than some other walks or there is a unique view that you see while walking, and there are obviously infinite many walks, each with probability 1/inf.

Now, there are 1/2 chance that you'll end up walking an "impossible" walk :D

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Is 0.00000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000.......................00000000000000 0000000001 equal to 0?

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If there are infinitely many zeros? Yes. Of course it is, for the same reason that .999...=1. Here's an intuitive argument:

Suppose that the number you have exists (and it does because completeness of the reals and it's lower bounded). Call it x. If there are infinitely many zeros, 10x also has inf. zeros before the one. So 10x=x implies x=0.

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if i am not mistaken this is a limit question,

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No!!

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Obviously not. It takes much more space :D

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If probability of some event occuring is \(\lim_{x\to\infty}\frac{1}{x}\) then can we say that it is almost impossible to happen?

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"If probability of some event occurring is lim 1/x as x -> infinity" is equivalent to saying "If probability of some event occurring is 0". Yes. The event is impossible.

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