Independence

Lesson 8 Chapter 2

We previously discussed conditional probabilities. You learn the concept of probability of an event $A$ dependent to another event $B$ as $P(A|B)$ . Intuitive, we can infer that if $A$ and $B$ are independent, then $P(A|B)$ does not depend on $B$ , and it simply equals to $P(A)$ . We previously investigated the formulation as well. Technically, being independent is mutual, i.e., when $P(A|B)=P(A)$ , then $P(B|A)=P(B)$ and vice verse. We can conclude that when $A$ and $B$ are independent, then $P(A \cap B)=P(A)P(B)$ . This formulation leads to the following definition:

Independent Events

Two events $A$ and $B$ are said to be independent if $P(A \cap B)=P(A)P(B)$ . Otherwise, they are called dependent.

Example

Assume we toss two dices. Let $\mathcal{E}$ denote the event that the sum of the dice is 5 and $\mathcal{P}$ indicate the event that the first dice equals 3. Can you confirm if the two events are independent or not?

Solution: To find the answer, we should calculate $P(\mathcal{E} \cap \mathcal{F})$ and $P(\mathcal{E})P(\mathcal{F})$ . If both are equal, then the two events are independent. Otherwise, they are dependent.

$\begin{equation*}\begin{split}& P(\mathcal{E} \cap \mathcal{F}) = P(\{(3,2)\}) = \frac{1}{36} \\& P(\mathcal{E})P(\mathcal{F}) = (\frac{1}{9})(\frac{1}{6}) = \frac{1}{54}\end{split}\end{equation*}$

How we did the calculations? Calculating $P(\mathcal{F})$ is simple for only one dice, the probability of getting any number is $\frac{1}{6}$ . But how are we going to calculate $P(\mathcal{E})$ ? The event space of $E$ is the following:

$E = \{(1,4),(2,3),(3,2),(4,1)\}$

Henceforth, $P(\mathcal{E})$ equals to $\frac{4}{36}=\frac{1}{9}$ . Basically, $\mathcal{E}$ represents all the possible situations that the sum of two dices equals to five. Since $P(\mathcal{E} \cap \mathcal{F}) \neq P(\mathcal{E})P(\mathcal{F})$ , then we can conclude that $\mathcal{E}$ and $\mathcal{F}$ are NOT independent.

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Independence

Independent Events

Example

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