Learning goals
By the end of this section, students should be able to translate ordinary language into events, compute probabilities using the axioms, recognize conditional probability and independence, and apply total probability and Bayes' theorem to real examples.
Events as sets
Use union, intersection, complement, and difference to describe probability statements.
Rules
Apply $P(A^c)=1-P(A)$ and $P(A\cup B)=P(A)+P(B)-P(A\cap B)$.
Updating
Use Bayes' theorem to update prior probability after observing evidence.
1. Sample space and events
A random experiment has a sample space $\Omega$, the set of all possible outcomes. An event is a subset of $\Omega$.
Example: two coin tosses
$$\Omega=\{HH,HT,TH,TT\}.$$
Let $A=$ “first toss is Head”, $B=$ “second toss is Head”, and $C=$ “the two tosses are the same”. Then
$$A=\{HH,HT\},\quad B=\{HH,TH\},\quad C=\{HH,TT\}.$$
Language to set notation
| Phrase | Event notation |
|---|---|
| not $A$ | $A^c$ |
| $A$ and $B$ | $A\cap B$ |
| $A$ or $B$ or both | $A\cup B$ |
| $A$ but not $B$ | $A\cap B^c$ |
| exactly one of $A,B$ | $(A\cap B^c)\cup(A^c\cap B)$ |
2. Probability axioms
Immediate consequences
- $P(\emptyset)=0$.
- $P(A^c)=1-P(A)$.
- If $A\subseteq B$, then $P(A)\le P(B)$.
- $P(A\cup B)=P(A)+P(B)-P(A\cap B)$.
Core warning: do not add P(A)+P(B) unless A and B are disjoint. The overlap A ∩ B is counted twice and must be subtracted once.3. Interactive event algebra lab
Move the sliders. The page checks whether the chosen values are valid and computes common event probabilities.
| Event | Probability |
|---|---|
| $A\cup B$ | |
| exactly one of $A,B$ | |
| at most one of $A,B$ | |
| neither $A$ nor $B$ |
4. Conditional probability
When $P(B)>0$, the conditional probability of $A$ given $B$ is $$P(A\mid B)=\frac{P(A\cap B)}{P(B)}.$$
Interpretation
Conditioning changes the sample space from $\Omega$ to $B$. Within the smaller world where $B$ has happened, we ask how much of $B$ is also in $A$.
Multiplication rule
Rearranging the definition gives
$$P(A\cap B)=P(A\mid B)P(B)=P(B\mid A)P(A).$$
5. Independence checker
Events $A$ and $B$ are independent if observing one does not change the probability of the other:
$$P(A\cap B)=P(A)P(B).$$
Two-coin example
Let $A=$ “first toss is Head” and $C=$ “the two tosses are the same”. Then $P(A)=1/2$, $P(C)=1/2$, and $P(A\cap C)=P(\{HH\})=1/4$. Since $1/4=(1/2)(1/2)$, the events are independent.
6. Law of total probability
If $B_1,\ldots,B_k$ form a partition of the sample space, then
$$P(A)=\sum_{i=1}^k P(A\mid B_i)P(B_i).$$
Medical test simulator
| Quantity | Value |
|---|---|
| $P(+)$ | |
| $P(D|+)$ |
Teaching point
A positive result can still have a modest posterior probability when the base rate is low. This is the base-rate effect.
For the default values, $P(+)=0.95(0.01)+0.10(0.99)=0.1085$.
7. Bayes' theorem
Bayes' theorem follows from applying the multiplication rule in two ways:
$$P(B_j\mid A)=\frac{P(A\mid B_j)P(B_j)}{\sum_i P(A\mid B_i)P(B_i)}.$$
Factory machine example
| Quantity | Value |
|---|---|
| $P(D)$ | |
| $P(I|D)$ |
Default calculation
With $P(I)=0.4$, $P(II)=0.6$, $P(D|I)=0.01$, and $P(D|II)=0.02$,
$$P(I|D)=\frac{0.01\cdot0.4}{0.01\cdot0.4+0.02\cdot0.6}=0.25.$$
8. Birthday paradox
Ignoring leap years, the probability that at least two people in a group of size $n$ share a birthday is
$$1-\frac{365\cdot364\cdots(365-n+1)}{365^n}.$$
| Probability | Value |
|---|---|
| all birthdays different | |
| at least one shared birthday |
Try $n=23$ and $n=30$. The result grows faster than intuition expects.
9. Geometric probability
For a point $(X,Y)$ chosen uniformly from the unit square, probability equals area. For example,
$$P(X^2+Y\le 1)=\int_0^1(1-x^2)\,dx=\frac{2}{3}.$$
Generalize the curve
Move $a$ in $Y\le 1-X^a$. The probability is
$$\int_0^1(1-x^a)\,dx=\frac{a}{a+1}.$$
Area probability:
Summary map
Set operations
$A^c$, $A\cup B$, $A\cap B$, $A\setminus B$ translate words into math.
Conditioning
$P(A|B)=P(A\cap B)/P(B)$ means “restrict the world to $B$.”
Updating
Bayes turns $P(E|H)$ and $P(H)$ into $P(H|E)$.
| Formula | When to use |
|---|---|
| $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ | “A or B” with possible overlap |
| $P(A^c)=1-P(A)$ | Use complements, especially for “at least one” |
| $P(A\cap B)=P(A|B)P(B)$ | Sequential or conditional information |
| $P(A)=\sum_iP(A|B_i)P(B_i)$ | Break into cases |
| $P(B_j|A)=\frac{P(A|B_j)P(B_j)}{\sum_iP(A|B_i)P(B_i)}$ | Reverse conditional probabilities |
Self-check quiz
These are designed for quick in-class checks. Students should explain why the chosen answer is correct.