# Complement of independent events are independent

• MHB
• mathmari
In summary, the conversation discusses the independence of events and how to show that their complements are also independent. The conversation includes a step-by-step calculation and checks for correctness, and also considers the concept of pairwise and mutual independence. The final expression is found, with a clarification about a small error in the original calculation.
mathmari
Gold Member
MHB
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)

mathmari said:
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 $P(A^c)P(B^c)P(C^c)$.

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)

## What is the definition of the complement of independent events?

The complement of independent events refers to the probability of the events not occurring together. In other words, it is the probability of one event occurring while the other does not.

## How do you determine if the complement of two events are independent?

To determine if the complement of two events are independent, you can use the formula P(A and B) = P(A) * P(B), where P(A) is the probability of event A and P(B) is the probability of event B. If the result is equal to the complement of the two events, then they are independent.

## Can the complement of two independent events be dependent?

No, the complement of two independent events cannot be dependent. If two events are independent, it means that the occurrence of one event does not affect the probability of the other event occurring. Therefore, their complements will also be independent.

## What is an example of two events that are independent?

An example of two independent events would be rolling a dice and flipping a coin. The outcome of rolling the dice does not affect the outcome of flipping the coin, making them independent events.

## How is the complement of independent events useful in probability calculations?

The complement of independent events is useful in calculating the probability of certain events not occurring together. It can also be used to calculate conditional probabilities, where the probability of one event occurring is dependent on the occurrence of the other event not occurring.

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