Math
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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.

Pen