Independence

Two events are classified as independent if the occurrence of one has no effect on the occurrence of the other. Given events \(A\) and \(B\), if they are independent, then \(P(A\cap B) = P(A) \cdot P(B)\)

Thus,

\[ P(A|B) = P(A) \\ P(B|A) = P(B) \]

Proof

We can use conditional probability to prove this. The event \(A\) is independent if its conditional probability is equal to \(P(A)\).

\[ P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A)\cdot P(B)}{P(B)} = P(A)\\ P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{P(A)\cdot P(B)}{P(A)} = P(B) \]

So, when events \(A\) and \(B\) are independent, knowing that \(B\) occurred does not change the probability of \(B\), and vice versa.

Let say we want to toss die and coin and calculate the probability of getting both a 5 on the die and a head on the coin. The result of the coin toss does not influence the outcome of the die roll, and vice versa.

The sample spaces for each event are:

\[ \Omega_A = \{H, T\}\\ \Omega_B = \{1, 2, 3, 4, 5, 6 \} \]

The probability of each event is:

\[ P(A) = \frac{1}{2} \\ P(B) = \frac{1}{6} \]

The probability of both events happening is:

\[ P(A\cap B) = \frac{1}{2} \cdot \frac{1}{6} = \frac{1}{12} \]