An insurance company needs $P(\text{flood AND car theft})$. A doctor needs $P(\text{disease} \mid \text{positive test})$. These require combining events. In this lesson you'll master the addition rule for "or" and the multiplication rule for "and", the two engines behind almost every probability calculation you'll meet in the HSC.
Today's hook, With $P(A) = 0.6$ and $P(B) = 0.4$, can $P(A \cup B)$ ever be less than $P(A)$? This sounds obvious, but getting the logic exactly right is what separates Band 4 from Band 6 answers. By the end of this lesson you'll know the proof cold.
0/5QUESTS
1
You’re here
Orient and recall
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
Worksheets
Practise this lesson
Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.
Consider two events $A$ and $B$ with $P(A) = 0.6$ and $P(B) = 0.4$. Without a formula, can $P(A \cup B)$ ever be less than $P(A)$? Explain your reasoning before reading on.
auto-saved
2
You’re here
The two moves
Work through the core explanation before applying it.
02
The two moves
+5 XP to read
This lesson is built on two rules. The first handles "or" problems; the second handles "and" problems. Lock both into memory and you can solve almost any two-event probability question.
Every combined-event question routes through one of two roads: addition rule when events are joined by "or" ($\cup$), or multiplication rule when events are joined by "and" ($\cap$).
$P(A \cap B) = P(A) \times P(B \mid A)$
Addition rule (OR)
$P(A \cup B) = P(A)+P(B)-P(A \cap B)$. Subtract intersection always even if the question doesn't mention it.
Multiplication rule (AND)
$P(A \cap B) = P(A) \times P(B \mid A)$. If independent, $P(B \mid A)= P(B)$ so it simplifies to $P(A) \times P(B)$.
Complement for "at least one"
$P(\text{at least one}) = 1 - P(\text{none})$. This is nearly always faster than direct counting.
3
You’re here
What you'll master
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
03
What you'll master
Know
Key facts
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$P(A \cap B) = P(A) \times P(B \mid A)$
$P(A') = 1 - P(A)$ and $P(\text{at least one}) = 1 - P(\text{none})$
Understand
Concepts
Why the addition rule subtracts the intersection once
How the multiplication rule builds probability through stages
The difference between mutually exclusive and independent events
Can do
Skills
Apply the addition rule to find $P(A \cup B)$ from a Venn diagram
Use the multiplication rule for two-stage experiments
Solve "at least one" problems efficiently with complements
04
Key terms
Addition rule$P(A \cup B) = P(A)+P(B)-P(A \cap B)$, prevents double-counting the overlap.
Multiplication rule$P(A \cap B) = P(A) \times P(B \mid A)$, chains two event probabilities together.
Conditional probability$P(B \mid A)$, the probability of $B$ given that $A$ has already occurred.
Independent events$A$ does not affect $B$: $P(B \mid A)=P(B)$, so $P(A \cap B) = P(A) \times P(B)$.
Mutually exclusive$P(A \cap B)=0$, so $P(A \cup B)=P(A)+P(B)$.
Complement$P(A') = 1-P(A)$; for "at least one": $1-P(\text{none})$.
4
You’re here
The addition rule for $P(A \cup B)$
Work through the core explanation before applying it.
05
The addition rule for $P(A \cup B)$
core concept
When we want the probability that at least one of two events occurs, we must not double-count outcomes that belong to both.
Shading both circles covers the overlap twice. We subtract $P(A \cap B)$ so it counts exactly once.
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
Bounds on $P(A \cup B)$: Since $P(A \cap B) \geq 0$, we always have $P(A \cup B) \leq P(A) + P(B)$. Also, since $A \subseteq (A \cup B)$, we have $P(A \cup B) \geq P(A)$. The union can never be smaller than either individual probability, which answers our Think First question.
Insurance risk. An insurer calculates $P(\text{flood} \cup \text{fire})$. They can't simply add the two probabilities because some properties are prone to both. The intersection, though small, must be subtracted to avoid overestimating risk and overcharging premiums.
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$, always subtract the intersection; $P(A \cup B) \geq \max(P(A), P(B))$, the union is always at least as large as either event
Pause, copy the addition rule $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ and the reason you subtract the intersection (to avoid double-counting the overlap) into your book.
Quick check: If $P(A) = 0.5$, $P(B) = 0.4$, and $P(A \cap B) = 0.2$, what is $P(A \cup B)$?
06
The multiplication rule for $P(A \cap B)$
core concept
We just saw that $P(A \cup B)$ uses addition with a correction for double-counting. That raises a question: what rule do we use for the intersection $P(A \cap B)$, the AND probability, especially when one event can influence the other? This card answers it → the multiplication rule: $P(A \cap B) = P(A) \times P(B \mid A)$, which simplifies to $P(A) \times P(B)$ only when events are independent.
When we want the probability that both events occur, we multiply probabilities, but we must account for how the first event affects the second.
$$P(A \cap B) = P(A) \times P(B \mid A)$$
Here $P(B \mid A)$ means "the probability of $B$ given that $A$ has occurred." This is conditional probability.
Example: A bag has 5 red and 5 blue marbles. You draw two without replacement. What is $P(\text{red then blue})$?
$P(\text{first red}) = \dfrac{5}{10} = \dfrac{1}{2}$. After removing one red, $P(\text{second blue} \mid \text{first red}) = \dfrac{5}{9}$.
So $P(\text{red then blue}) = \dfrac{1}{2} \times \dfrac{5}{9} = \dfrac{5}{18}$.
Independent events special case: If $A$ does not affect $B$, then $P(B \mid A) = P(B)$ and:
$$P(A \cap B) = P(A) \times P(B) \quad \text{(when independent)}$$
General: $P(A \cap B) = P(A) \times P(B \mid A)$, always check whether events affect each other; Independent: $P(B \mid A) = P(B)$, so $P(A \cap B) = P(A) \times P(B)$
Pause, copy both multiplication forms: general $P(A \cap B) = P(A) \times P(B \mid A)$ and the simpler independent version $P(A \cap B) = P(A) \times P(B)$, only use the simple version when independence is confirmed into your book.
Did you get this? True or false: if $A$ and $B$ are independent, then $P(A \cap B) = P(A) \times P(B)$.
Worked examples · 3 in a row, reveal as you go
5
You’re here
Work examples end to end
Follow the reasoning through complete worked solutions.
PROBLEM 1 · ADDITION RULE
If $P(A) = 0.6$, $P(B) = 0.5$, and $P(A \cap B) = 0.3$, find: (a) $P(A \cup B)$; (b) $P(A' \cap B)$.
1
$P(A \cup B) = 0.6 + 0.5 - 0.3 = 0.8$
Apply the addition rule. Subtract the intersection once.
2
$P(A' \cap B) = P(B) - P(A \cap B) = 0.5 - 0.3 = 0.2$
"$B$ but not $A$" = all of $B$ minus the part shared with $A$. Verify: $P(A \cup B) + P(A' \cap B') = 0.8 + (1-0.8) = 1$ ✓
PROBLEM 2 · MULTIPLICATION RULE
A factory: Machine A makes 60% of items with 2% defect rate; Machine B makes 40% with 5% defect rate. Find: (a) $P(\text{defective})$; (b) $P(\text{from A and defective})$.
Each roll is independent. Use the complement: $P(\text{no six})$ on each roll is $\frac{5}{6}$.
2
$P(\text{at least one 6}) = 1 - \dfrac{125}{216} = \dfrac{91}{216} \approx 0.421$
Subtract from 1. Direct counting would need: exactly one 6, exactly two 6s, exactly three 6s, three separate calculations. Complement is one calculation.
Fill the gap: A box contains 6 red and 4 green balls. Two are drawn without replacement. $P(\text{red then green}) = \dfrac{6}{10} \times \dfrac{4}{9} = $ .
6
You’re here
The power of complementary events
Work through the core explanation before applying it.
10
The power of complementary events
core concept
We just saw that both the addition and multiplication rules require you to check whether events are mutually exclusive or independent before applying the simplified form. That raises a question: what is the most efficient way to confirm those conditions and avoid using the wrong simplification? This card answers it → use the definitions as tests: mutually exclusive means $P(A \cap B) = 0$; independent means $P(A \cap B) = P(A) \times P(B)$.
The complement rule is deceptively simple but extraordinarily powerful:
$$P(\text{at least one}) = 1 - P(\text{none})$$
Its most common application is "at least one" problems. Directly counting "at least one success" requires summing probabilities for exactly 1, exactly 2, exactly 3, and so on. The complement needs only one calculation.
Example: Four people each guess a number from 1–10. What is $P(\text{at least one correct})$?
So $P(\text{at least one correct}) = 1 - 0.6561 = 0.3439$.
Misconception alert. $P(A \cup B) = P(A) + P(B)$ only works when $A$ and $B$ are mutually exclusive. $P(A \cap B) = P(A) \times P(B)$ only works when $A$ and $B$ are independent. Using the simple product when events are dependent (e.g. drawing without replacement) is one of the most common HSC errors.
$P(A \cup B) = P(A)+P(B)$ only if mutually exclusive (given or proven); $P(A \cap B) = P(A) \times P(B)$ only if independent (given or proven)
Pause, copy both condition tests: mutually exclusive requires $P(A \cap B) = 0$ (given or proven); independent requires $P(A \cap B) = P(A) \times P(B)$ (given or proven, never assume!) into your book.
Misconceptions to fix · the 3 traps that cost marks
7
You’re here
Dodge the traps, then apply
Meet the mistakes that cost marks, then do it yourself.
Trap 01
Using simple addition for non-exclusive events
Writing $P(A \cup B) = P(A) + P(B)$ without checking if events are mutually exclusive. This only works when $P(A \cap B) = 0$. Otherwise you must subtract the intersection.
Trap 02
Using simple product for dependent events
Writing $P(A \cap B) = P(A) \times P(B)$ when drawing without replacement. After removing the first item, the denominator shrinks, the events are dependent. Use $P(A) \times P(B \mid A)$ instead.
Trap 03
Trying to count "at least one" directly
Listing all the "at least one" cases takes multiple calculations and risks missing a case. Always flip to the complement: $1 - P(\text{none})$, one clean calculation.
Did you get this? True or false: $P(A \cap B) = P(A) \times P(B)$ is always true for any two events $A$ and $B$.
8
You’re here
Build the idea
Work through the core explanation before applying it.
Activities · practice with the ideas
1
If $P(A) = 0.5$, $P(B) = 0.4$, and $P(A \cap B) = 0.2$, find $P(A \cup B)$.
2
A box contains 6 red and 4 green balls. Two are drawn without replacement. Find $P(\text{red then green})$.
3
If $P(A) = 0.3$, $P(B) = 0.6$, and $A$ and $B$ are mutually exclusive, find $P(A \cup B)$.
4
A bag has 8 black and 12 white counters. Two are drawn with replacement. Find $P(\text{both black})$.
5
Prove that $P(A \cap B) \leq \min(P(A), P(B))$ and explain when equality holds.
12
Revisit your thinking
Earlier you were asked: can $P(A \cup B)$ ever be less than $P(A)$?
$P(A \cup B)$ cannot be less than $P(A)$. By definition, $A \subseteq (A \cup B)$, so every outcome in $A$ is also in $A \cup B$. Therefore $P(A \cup B) \geq P(A)$. Similarly, $P(A \cup B) \geq P(B)$. The smallest $P(A \cup B)$ can be is $\max(P(A), P(B))$, which occurs when one event is a subset of the other.
auto-saved
1
You’re here
Multiple choice
Answer the drill bank and rate your confidence.
01
Multiple choice
+5 XP per correct · +25 XP all-correct
Pick your answer, then rate your confidence, that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
2
You’re here
Short answer
Write full responses, then check them against the model answers.
02
Short answer
ApplyBand 43 marks
Q1. A bag contains 4 red, 5 blue, and 6 green marbles. Two marbles are drawn without replacement. Find: (a) $P(\text{both red})$; (b) $P(\text{red then blue})$; (c) $P(\text{at least one red})$. (3 marks)
auto-saved
ApplyBand 43 marks
Q2. In a survey, $P(A) = 0.6$, $P(B) = 0.5$, and $P(A \cap B) = 0.3$. (a) Find $P(A \cup B)$. (b) Find $P(A' \cap B)$, the probability that $B$ occurs but $A$ does not. (c) Verify that $P(A \cup B) + P(A' \cap B') = 1$. (3 marks)
auto-saved
AnalyseBand 53 marks
Q3. A fair coin is flipped until either two heads appear or four flips have been made, whichever comes first. (a) Draw a tree diagram showing all possible sequences. (b) Find the probability the experiment ends with two heads. (c) Explain why the multiplication rule must be used at each branch of the tree, and why simply counting paths gives the wrong answer. (3 marks)
auto-saved
Comprehensive answers (click to reveal)
Activity 1: 1. $0.5+0.4-0.2=0.7$ · 2. $\frac{6}{10}\times\frac{4}{9}=\frac{4}{15}$ · 3. $0.3+0.6=0.9$ (mut. excl.) · 4. $\frac{8}{20}\times\frac{8}{20}=\frac{4}{25}=0.16$ · 5. Since $A \cap B \subseteq A$, $n(A \cap B) \leq n(A)$ so $P(A \cap B) \leq P(A)$; similarly $\leq P(B)$. Equality when one event is a subset of the other.
Q3 (3 marks): (a) Paths: HH, HTH, HTTH, HTTT, THH, THTH, THTT, TTHH, TTHT, TTTH, TTTT [0.5]. (b) Ends-with-2H paths: HH($\frac{1}{4}$), HTH($\frac{1}{8}$), HTTH($\frac{1}{16}$), THH($\frac{1}{8}$), THTH($\frac{1}{16}$), TTHH($\frac{1}{16}$). Sum = $\frac{4+2+1+2+1+1}{16} = \frac{11}{16}$ [1]. (c) Each branch probability requires multiplication because the stopping condition changes at each stage. Paths are not equally likely (length varies), so counting them gives the wrong answer. The multiplication rule correctly weights each path [1.5].
1
You’re here
Review and finish
Take the module quiz if you are ready, then mark the lesson complete.
01
Take the full module quiz
quiz
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
