Proof that the derivative of the prime counting function is the probability of prime? The derivative of the estimation of the prime counting function, $\frac{x}{ln(x)}$, is $\frac{ln(x)-1}{ln(x)^2}$, which is approximately $\frac{1}{lnx}$ for large values of $x$. According to the prime number theory, $\frac{1}{lnx}$ is the probability that a randomly chosen integer between 2 and $x$ is a prime number.
Why is the derivative of the prime counting function the probability of getting a prime number?
Edit: I've only studied up to basic integrals, so try to keep it simple!

This is not precise, because $\pi(x)$ is neither continuous nor differentiable, but you get the idea :-)
Mean value theorem:
$$
\pi(x+1)-\pi(x)=\pi'(\xi)[(x+1)-x]=\pi'(\xi)
$$
for some $\xi\in(x,x+1)$. But $\pi(x+1)-\pi(x)$ is either zero or one according to whether $x+1$ is prime or not. So probabilistically we get the result you described. Or you can look at a longer interval. 
