Let's say we now have two stacks of powders, one is $a$ the other is $b$. The ratio of amount of $a:b$ is $p:(1-p)$.
The two stacks of powders are poured together (without mixing). It forms two compartments $A$ and $B$. Now we draw a sample of $n$ particles. We are interested in the proportion of $a:b$ drawn in sample.
Let $X$ be the number of particles successfully drawn as $a$. The book said $Var(\frac{X}{n})=p(1-p)$ (variance in proportion).
This is reasonable if we are limited in one compartment. If $A$ is picked then all $n$ particles are $a$, and same applies for $B$. But what if we pick samples near the boundary of $A$ and $B$?
My argument is $$ \begin{align*} Pr(X>0) &=Pr(X>0|A)Pr(A)+Pr(X>0|B)Pr(B)+Pr(X>0|boundary)Pr(boundary) \\ &=1+0+Pr(X>0|boundary)Pr(boundary) \end{align*}$$ and $Pr(X=n)\le1$ which forces $Pr(X>0|boundary)Pr(boundary)=0$.
Why is this term zero in physical meaning?
p.s. in the original question, the powders would then be well-mixed and the new variance is compared with the above one.