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Let A be an event such that P[A] ∈ (0, 1). Show that A and AC are not independent. Does this still hold if P[A] ∈ {0, 1}?

The solution is: We have P[A∩A^C] = P[∅] = 0. On the other hand, P[A]·P[A^C] = PA. The events are independent if, and only if, these two quantities are equal, i.e., if, and only if, P[A] = 0 or P[A] = 1.

Are the events independent, because for both P[A] = 0 or 1, P[A]·P[A^C] = PA = 0?

user10354138
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Rayleigh
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  • You have several errors and mistakes in typesetting this. For example you say "On the other hand, P[A]·P[A^C] = PA." What do you mean by PA? That does not match the notation you have been using till now. You should have found $\Pr(A)\cdot\Pr(A^c)=\Pr(A)\cdot(1-\Pr(A))$ which you should know is equal to zero iff $\Pr(A)$ is zero or $\Pr(A)=1$. – JMoravitz Oct 24 '23 at 13:32
  • The punchline is that mutually exclusivity is not the same as mutual independence, and that you can not have events be both except in the extreme cases of either (almost) impossible events (probability zero) or (almost) sure events (probability 1). – JMoravitz Oct 24 '23 at 13:34

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