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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?

hemant kumar
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    You are confusing "selecting independently at random two subsets of $S$ (from some probability distribution over all/some subsets of $S$") with "selecting two events $A,B\subseteq S$ which are independent as events in the uniform probability space on $S$". – Michal Adamaszek May 24 '23 at 11:34
  • To be clear... you say "suppose I have a set as a sample space..." but you did not clarify or define what the probability distribution is over that sample space. $P(B)=\dfrac{n(B)}{6}$ here is only guaranteed to be true if you were to explicitly state that you wanted the probability distribution here to be uniform. There are many other probability distributions that could have worked here where the probability of $B$ need not equal that. – JMoravitz May 24 '23 at 12:06
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    "can someone explain why choice of [[arbitrarily selected sets]] not independent?" Simply because certain events will be independent and other certain events will not be independent. To check if they are independent, you check the definition... which is the $P(A\cap B)=P(A)P(B)$ property alluded to before. That is the definition. Attempts to understand what constitutes independence through intuition can and will fail repeatedly. There are plenty of examples where one might have guessed things were independent for it to fail and not actually be independent, or vice versa. – JMoravitz May 24 '23 at 12:09

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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.

ryang
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  • ''A and B are independent precisely when B has equally many outcomes common with A and outcomes not common with A''. doesn't that mean $P(A\cap B) = P(A\cap \overline{B})$? Do you instead want to say, the distribution of B in sample space is same as its distribution in $A\cap B$ – hemant kumar May 24 '23 at 17:32
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    No and no. B's outcomes common with A and B's outcomes not common with A are equinumerous. $\quad$ P.S. I do encourage you to click on that link, as well as the link in that link; you may find the diagrams elucidating! – ryang May 24 '23 at 17:57