MHB Complement of independent events are independent

AI Thread Summary
The discussion focuses on proving that the complements of independent events, specifically $A^c, B^c, C^c$, are also independent. The participants analyze the probability expressions involving these events and confirm that the expression for $P(A^c \cap B^c \cap C^c)$ simplifies correctly to the product of their individual probabilities. There is a clarification regarding the evaluation of the union of events and the necessity of including the intersection of all three events in the probability calculations. The conversation concludes with one participant expressing satisfaction upon reaching the desired expression for independence. This indicates a successful understanding of the relationship between independent events and their complements.
mathmari
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Hey! :o

Let $A,B,C$ be independent events. I want to show that $A^c, B^c, C^c$ are independent.

We have the following: \begin{align*}&P(A^c\cap B^c\cap C^c)=1-P(A\cup B\cup C)\\ & =1-\left [P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)-P(A\cap B\cap C)\right ] \\ & = 1-\left [P(A)+P(B)+P(C)-P(A)P(B)-P(A)P(C)-P(B)P( C)-P(A)P (B)P( C)\right ] \\ & = 1-P(A)-P(B)-P(C)+P(A)P(B)+P(A)P(C)+P(B)P( C)+P(A)P (B)P( C) \\ & = 1+P(A)\left [P(B)-1\right ]+P(C)\left [P(A)-1\right ]+P(B)\left [P( C)-1\right ]+P(A)P (B)P( C) \\ & = 1-P(A)P(B^c)-P(C)P(A^c)-P(B)P( C^c)+P(A)P (B)P( C)\end{align*} Is everything correct so far? How could we continue? (Wondering)
 
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mathmari said:
Hey! :o

Let $A,B,C$ be independent events. I want to show that $A^c, B^c, C^c$ are independent.

We have the following: \begin{align*}&P(A^c\cap B^c\cap C^c)=1-P(A\cup B\cup C)\\ & =1-\left [P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)-P(A\cap B\cap C)\right ] \\ & = 1-\left [P(A)+P(B)+P(C)-P(A)P(B)-P(A)P(C)-P(B)P( C)-P(A)P (B)P( C)\right ] \\ & = 1-P(A)-P(B)-P(C)+P(A)P(B)+P(A)P(C)+P(B)P( C)+P(A)P (B)P( C) \\ & = 1+P(A)\left [P(B)-1\right ]+P(C)\left [P(A)-1\right ]+P(B)\left [P( C)-1\right ]+P(A)P (B)P( C) \\ & = 1-P(A)P(B^c)-P(C)P(A^c)-P(B)P( C^c)+P(A)P (B)P( C)\end{align*} Is everything correct so far? How could we continue? (Wondering)

Note that the expression is same as $(1-P(A))(1-P(B))(1-P(C))$, which is same as $P(A^c)P(B^c)P(C^c)$.
 
Hey mathmari! (Smile)

If $A,B,C$ are independent events, are they pairwise independent?
That is $P(A\cap B)=P(A)P(B)$, and the same for the other pairs?

Or are they mutually independent?

Anyway, let's start with $A^c,B^c$. These are independent iff $P(A^c\cap B^c)=P(A^c)P(B^c)$.
Can we evaluate both $P(A^c\cap B^c)$ and $P(A^c)P(B^c)$ to see if they are equal? (Wondering)
mathmari said:
$$1-P(A\cup B\cup C) =1-\left [P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)-P(A\cap B\cap C)\right ]$$

Shouldn't that be:
$$1-P(A\cup B\cup C) =1-\left [P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C){\color{red}+}P(A\cap B\cap C)\right ]$$
(Wondering)
 
Thank you for the hints! I got now the desired expression! (Yes)
 
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