Selecting Independent events from a Set

Question

Suppose I have a set as sample space $S = {\{1,2,3,4,5,6\}}$

Now I want to select two independent events $A$ and $B$ from this set.

Suppose I selected $\{1,2,3\}$ as set $A$. Since set B is independent of A, I should be free to choose anything as set B.

But, by definition of independent events $$P(A\cap B) = P(A)P(B)$$ $$\implies \frac{n(A\cap B)}6 = \frac{n(A)}6 \frac{n(B)}6$$ $$\implies n(A\cap B) = \frac{n(A)n(B)}6$$ It sets a condition of $A\cap B$, so we are not free to choose any arbitrary subset as B despite it being independent on A.

can someone explain why choice of independent sets not independent?

Answer 1

Suppose I have a set as sample space $S = {\{1,2,3,4,5,6\}}$ Suppose I selected $\{1,2,3\}$ as set $A$.

Since set B is independent of A, I should be free to choose anything as set B.

That's not what independence of events means.

Let event A have a nonzero probability. Then events $A$ and $B$ are independent iff knowing that $A$ happens doesn't change $B$'s probability (not "doesn't change how $B$ can happen" and not "doesn't change how many ways $B$ can happen").

But, by definition of independent events, $$n(A\cap B) = \frac{n(A)n(B)}6$$

Yes, in your particular classical-probability scenario, indeed, $A$ and $B$ are independent precisely when $B$ has equally many outcomes common with $A$ and outcomes not common with $A.$

so we are not free to choose anything as B despite it being independent on A.

You mean that not every subset can be chosen as $B$ (rather that that no subset can be chosen as $B$); this is correct.