# 14 - Independent events, Annie, Boris, and Charlie

## Independent events
**Definition:** Two events $A$ and $B$ are **independent** if $Pr(A \cap B) = Pr(A) * Pr(B)$

**Consequence:** If $A$ and $B$ are independent and $Pr(B) > 0$ then $Pr(A | B) = \frac{Pr(A\cap B)}{Pr(B)} = \frac{Pr(A)Pr(B)}{Pr(B)} = Pr(A)$

Ex: Rolling two dice

$S = \{(D_1, D_2): D_1, D_2 \in \{1,2,3,4,5,6\}\}$ $Pr(\omega ) = \frac{1}{|S|} = \frac{1}{36}$

$A = D_2 = 6 = \{(1,6),(2,6),(3,6),(4,6),(5,6),(6,6)\}$
$B = D_1 = 4 = \{(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\}$

$Pr(A) = \frac{|A|}{|S|} = \frac{6}{36} = \frac{1}{6}$
$Pr(B) = \frac{|B|}{|S|} = \frac{6}{36} = \frac{1}{6}$

Another ex:

$A = D_1+D_2=7 = \{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\}$
$B = D_1 =4 = \{(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\}$

$Pr(A) = \frac{|A|}{|S|} = \frac{6}{36} = \frac{1}{6}$
$Pr(B) = \frac{|B|}{|S|} = \frac{6}{36} = \frac{1}{6}$

$Pr(A \cap B) = Pr(\{(4,3)\}) = \frac{1}{36} = \frac{1}{6} * \frac{1}{6} = Pr(A) \times Pr(B)$

$\therefore$ A and B are independent!

### Base property
If $A$ and $B$ are independent then $A$ and $\bar{B}$ are independent

To prove this we need to show that $Pr(A \cap \bar{B}) = Pr(A) \times Pr(\bar{B}) = Pr(A) \times (1 - Pr(B))$

Because $Pr(A \cap \bar{B})$ and $Pr(A \cap B)$ are disjoint, then
$Pr(A) = Pr(A \cap \bar{B}) + Pr(A \cap B)$
$Pr(A) = Pr(A) * Pr(B) + Pr(A \cap \bar{B})$
$Pr(A \cap \bar{B}) = Pr(A) - Pr(A) * Pr(B)$
$= Pr(A) * (1 - Pr(B)) = Pr(A) * Pr(\bar{B})$

## Multi independence
Pairwise independent: A sequence of events $A_1, A_2,..., A_n$ is **Pairwise independance** if for any $1 \leq i \leq j \leq n$, $Pr(A_i \cap A_j) = Pr(A_i) \times Pr(A_j)$
In other words, $A_i$ and $A_j$ are independent

Mutually independent: A sequence of events $A_1, A_2,..., A_n$ is **mutually independent** if for each $k \in \{2,3,4,...,n\}$ and each $1 \leq i_1 \leq i_2 \leq ... \leq i_k \leq n$ (for each subset of events of size $\geq 2$)
$Pr(A_{i_1} \cap A_{i_2} \cap ... \cap A_{i_n}) = Pr(A_{i_1}) * Pr(A_{i_2}) * ... * Pr(A_{i_n})$

Example where these two are different:

$S = \{00, 01, 10, 11\}$ $Pr(\omega ) = \frac{1}{4}$ Bitstring $b_1, b_2$

$A = b_1 = 1 = \{10, 11\}$
$B = b_2 = 1 = \{01, 11\}$
$C = b_1 \neq b_2 = \{10, 01\}$

$Pr(A) = \frac{2}{4} = \frac{1}{2} = Pr(B) = Pr(C)$

Check for pairwise independence
$Pr(A \cap B) = Pr(\{11\}) = \frac{1}{4} = \frac{1}{2} \times \frac{1}{2} = Pr(A) \times Pr(B)$
$Pr(A \cap C) = Pr(\{10\}) = \frac{1}{4} = \frac{1}{2} \times \frac{1}{2} = Pr(A) \times Pr(C)$
$Pr(B \cap C) = Pr(\{01\}) = \frac{1}{4} = \frac{1}{2} \times \frac{1}{2} = Pr(B) \times Pr(C)$

Therefore $A, B$ and $C$ are pairwise independent.

Now to check for mutual independence
$Pr(A \cap B \cap C) = Pr(\{\emptyset \}) = 0 \neq \frac{1}{8} = Pr(A) \times Pr(B) \times Pr(C)$

Therefore $A, B$ and $C$ are not mutually independent.

## 3 Students
Alina, Bart and Charlie
3 Students, each student writes a 1-question all or nothing exam.
$Pr(\text{Alina answers correctly}) = Pr(A) = \frac{9}{10}$
$Pr(B) = \frac{9}{10}$
$Pr(C) = \frac{6}{10}$

$A, B, C$ are mutually independent.

$E = \text{At least 2 students pass / answer correctly} = (A \cap B) \cup (B \cap C) \cup (A \cap C)$
$Pr(E) = Pr(A \cap B) + Pr(B \cap C) + Pr(A \cap C) - 2Pr(A \cap B \cap C)$
$Pr(E) = (Pr(A) * Pr(B)) + (Pr(C) * Pr(B)) + (Pr(A) * Pr(C)) - 2(Pr(A) * Pr(B)* Pr(C))$
$= (\frac{9}{10} * \frac{9}{10}) + (\frac{9}{10} * \frac{6}{10}) + (\frac{9}{10} * \frac{6}{10}) - 2(\frac{9}{10} * \frac{9}{10} * \frac{6}{10}) = \frac{459}{500} = 0.918$

**Second scenario**
Charlie copies Barts answer, now 2 students pass if Bart passes

$E = B$
$E = Pr(B) = \frac{9}{10} = 0.90$

###### tags: `COMP2804`, `Probability`