I am reading about smooth numbers to find the time-complexity of the quadratic sieve (as a side topic on a cryptography book), and the author makes the following claim:
The probability that a randomly chosen number modulo $N$ is $B$-smooth is $\frac{\psi(N,B)}{N}$. In order to find $\pi(B)$ numbers that are B-smooth, we need to check approximately:
$\frac{\pi(B)}{\psi(N,B)/N}$ numbers
where $\pi(B)$ is the number of primes less than or equal to $B$, $\psi(N,B)$ is the number of $B$-smooth numbers less than $N$.
The question is I don't know how the author got to this approximate value.
Maybe the author is computing the expectation of the number of times before $m$ B-smooth numbers are found. Let $T_i$ be the number of samples we draw between the $i-1^{th}$ success and the $i^{th}$ success. Then we are interested in:
$$\mathbb E [\sum_i^m T_i] = \sum_i^m \mathbb E [T_i] = m * 1/p$$
where the first equality follows by linearity of expectation, and the second comes from the fact that $T_i$ each follows a geometric distribution with parameter $p$.
In your problem, $m$ is $\pi(B)$ and $p$ is $\psi(B, N)/N$