NotesAtHKU

Probability notation and event types


A and B	A or B	Not A
Intersection	Union	Compliment

A and B mutually exclusive $\leftrightarrow P(A\cap B) = 0\ \|\|\ P(A\cup B) = P(A) + P(B)$ .

This means that the events cannot happen at the same time.

A and B collectively exhaustive $\leftrightarrow P(A\cup B) = 1$

This means that one of the events must happen.

The probability of A given that B occurs: $P(A\ \|\ B) = \frac{P(A\cup B)}{P(B)}$

Bayes' theorem: $P(A\|B) = \frac{P(B\|A) \cdot P(A)}{P(B)} = \frac{P(A \cap B)}{P(B\|A) \cdot P(A) + P(B\|A') \cdot P(A')}$

A and B independent $\leftrightarrow (A) \times P(B) = P(A\cap B)$ .

Being independent means that the probability of an event has no influence on the other

Example

Two dice R and B are rolled.

For question 1:

\begin{align*} P(A)\times P(C) &= \frac{1}{36}\\ P(A\cap C) = P(6,1) &= \frac{1}{36},\quad\text{as R must be 6 and B must be 1}\\ \therefore P(A)\times P(C) &= P(A\cap C)\implies A\ \text{and}\ C\ \text{are independent} \end{align*}

For question 2:

\begin{align*} P(B)\times P(C) &= \frac{1}{216}\\ P(B\cap C) = P(6,6) &= \frac{1}{36}\\ \therefore P(B)\times P(C) &\neq P(B\cap C)\implies B\ \text{and}\ C\ \text{are not independent} \end{align*}