What is the probability of selecting 0 from the [0, 1] set? I'm a high school student so don't get mad at me for asking this question.
So if $p(i)$ equals to probability of getting $i$ as outcome then we have:
1 - For every $i$: $p(i)=p$
2 - $lim_{n\to\infty} (np)=1$
So p is exactly 0. Because if p is some real number greater than zero then $pn$ can become infinitely large.
But it contradicts with the fact that $lim_{n\to\infty}(pn)=1$! because $0.n=0$ not $1$!
Thanks in advance :).
 A: Contrary to what was expressed in the comments you don't need to know measure theory to understand what's going on here (though it would help, is highly encouraged, and is more rigorous and general than what I will outline below).
Instead you can think of the uniform distribution on $[0,1]$ as an approximation to more well-understood uniform distributions on finite sets. Let's start with a random variable that has an equal probability of being $0.05,0.15,0.25,\ldots 0.95.$ Since there are ten possible values, uniformity gives that the probability of each must be $1/10.$ That's not a uniform distribution on $[0,1]$ but it's an approximation of sorts. 
To make a better approximation, just add more points. Instead, we can do twenty points at $0.025,.0075,0.0125,\ldots,0975.$ We can give each of these twenty points $1/20$ probability for a uniform distribution. Clearly we can generalize this to $n$ points. where we have the allowed values $\frac{i-1/2}{n}$ for $i=1$ to $n$,  with $1/n$ probability each.
First, notice that as $n$ increases, the probability of taking any particular value decreases, limiting to zero as $n\to\infty.$ However for any small interval $(a,b)$ about that value, the probability of the number lying somewhere within that interval will be finite for large enough $n$.
If you work through this carefully you will see that the probability of the value lying in an interval $(a,b)$ is just $(b-a)$ (provided of course that the interval lies within $[0,1]).$ Also, it doesn't matter whether the interval is open, closed, or half open, which makes sense since the probability of being at either of the endpoints goes to zero as $n\to \infty.$
If this reminds you at all of Riemann sums you are beginning to see the connection to integration theory.
A: Assuming a continuous (smooth) probability distribution, the probability of getting exactly $x$ for any $x\in[0,1]$ is zero.  I know that sounds nuts, but that's how it works.  You have to ask what is the chance of getting a value in some little piece of $[0,1]$.
A: You can think of it like
$$\lim_{n\to ∞}p=\frac{1}{n}$$
so that
$$\lim_{n\to ∞}(np)=n\frac{1}{n}=1$$
What goes weird in your calculations is that 1) you assume multiplying any variable with a variable that approaches to infinity results in infinity which is not always true, 2) $p$ is infinitsimally small, $n$ (as you define it by limit) is infinitely large, so it is possible that their multiplication will result in a constant number; it could also be 0 or infinity.
Instead of focusing on algebraic side of limits on this issue, better focus on the fact that for any single value, its probability is a "line" whereas the total probability is an "area" and the area of that "area" is 1, whereas the area of a "line" is infinitsimally small (approaches to zero) and the number of numbers on the x-axis is infinitely large.
I suggest you reading this as another example that will help your pain go away:
https://en.wikipedia.org/wiki/Dirac_delta_function
