Probability and with replacement sampling

Question

I'm reading the book "Model Assisted Survey Sampling" from Särndal et al. In chapter 2, there's a section about Sampling with replacement. I'll put this into context: We have $m$ independent draws, such that, in every draw, every one of the $N$ population elements has the same selection probability : $\frac{1}{N}$

Once drawn, an element is replaced into the population so that all $N$ elements participate in each draw. Obviusly, the probability that any given element is not drawn at all is given by: $(1 - \frac{1}{N})^m$

So, the first order inclusion probability is: $\pi_k = 1 - (1- \frac{1}{N})^m$

Now, my particular cuestion is why is the second order inclusion probability is:

$\pi_{kl} = 1 - 2(1- \frac{1}{N})^m + (1- \frac{2}{N})^m$

I really don't understand why. Does this suppose to mean that $2(1- \frac{1}{N})^m - (1- \frac{2}{N})^m$ is the probability that neither the observation $k$ nor $l$ are drawn in the $m$ draws?

Please, if someone has an intuitive explanation, i would be very thankfull.

Answer 1

$\pi_{kl}$ is the probability that both $k$ and $l$ are included in the sample . Using $$P(AB)=P(A)+P(B)-P(A\cup B)$$, this is the prob that $k$ is included plus the prob that $l$ is included (these are both equal to $1-(1-\frac1N)^m$) minus the prob that at least one is included (this is equal to $1-(1-\frac2N)^m$).

Answer 2

Let $k\ne l$. The expression $$2\left(1-\frac{1}{N}\right)^m -\left(1-\frac{2}{N}\right)^m$$ is the probability that at least one of $k$ and $l$ is missing. So the expression of the post gives the probability that both are present.