Complement of independent events are independent

[mathmari]

Hey!

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)

 

[caffeinemachine]

Hey!

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 $P(A^c)P(B^c)P(C^c)$.

 

[I like Serena]

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)

$$1-P(A\cup B\cup C) =1-\left [P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)-P(A\cap B\cap C)\right ]$$

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)

 

[mathmari]

Thank you for the hints! I got now the desired expression! (Yes)