# 4.3 Independence & Causality

## Labeled Balls

I have four balls in an urn, each one distinctly labeled with either 110, 101, 011, or 000. I draw one at random.

Let $$A_1$$ denote the event that I draw a ball with a 1 in the $$1^{st}$$ position. Let $$A_2$$ and $$A_3$$ be defined similarly.

Which of the following are true about $$A_1,A_2,A_3$$?

Exercise 1

First note that they all have equal probability, because \begin{align}\Pr[A_1] &= \Pr[110\cup 101]= \frac{1}{2}&\\ \Pr[A_2] &= \Pr[110\cup 011]= \frac{1}{2}&\\ \Pr[A_3] &= \Pr[101\cup 011]= \frac{1}{2}.&\end{align}

They are also pairwise independent, since \begin{align}\Pr[A_1 \cap A_2]&= \Pr[110]=\frac{1}{4}=\frac{1}{2}\cdot\frac{1}{2}=\Pr[A_1]\Pr[A_2]&\\ \Pr[A_1 \cap A_3]&=\Pr[101]=\frac{1}{4}=\frac{1}{2}\cdot\frac{1}{2}=\Pr[A_1]\Pr[A_3]&\\ \Pr[A_2 \cap A_3]&=\Pr[011]=\frac{1}{4}=\frac{1}{2}\cdot\frac{1}{2}=\Pr[A_2]\Pr[A_3].&\end{align}

They are not mutually independent because $$\Pr[A_1\cap A_2\cap A_3] = \Pr[\emptyset] = 0 \neq \frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=Pr[A_1]Pr[A_2]Pr[A_3]$$.

They are not pairwise disjoint, since $$A_1\cap A_2=\{110\}.$$