Probability
- Use the probability scale from 0 to 1 and calculate relative frequency from experimental data
- Apply the AND rule $P(A \cap B) = P(A) \times P(B)$ for independent events
- Apply the OR rule $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ for any two events
- Draw and interpret Venn diagrams for 2 and 3 sets, including the inclusion–exclusion principle
- Find conditional probabilities using $P(A|B) = \dfrac{P(A \cap B)}{P(B)}$ and verify independence
🔑 Core Concepts
The Probability Scale
Probability quantifies how likely an event is to occur. Every probability is a number between 0 and 1 inclusive. A probability of 0 means the event is impossible; a probability of 1 means it is certain. Values in between describe the spectrum of likelihood.
Experimental vs Theoretical Probability
Theoretical probability is derived from the structure of the experiment (e.g., a fair six-sided die gives $P(4) = \tfrac{1}{6}$ by symmetry). Experimental probability (also called relative frequency) is estimated from data collected by actually performing the experiment.
Mutually Exclusive and Exhaustive Events
The AND Rule — Independent Events
Two events are independent if the occurrence of one has no effect on the probability of the other. When events are independent, you multiply their probabilities to find the probability that both occur.
The OR Rule — Addition Rule
The OR rule gives the probability that at least one of two events occurs. When both events can occur simultaneously (they are not mutually exclusive), simply adding the individual probabilities would double-count the overlap region. The intersection must be subtracted once to correct for this.
Venn Diagrams
Venn diagrams partition the sample space into distinct regions. Each region corresponds to a specific combination of events occurring or not occurring. The rectangle represents the entire sample space $S$.
- Only $A$ (in $A$ but not $B$): $P(A) - P(A \cap B)$
- $A \cap B$ (in both): $P(A \cap B)$
- Only $B$ (in $B$ but not $A$): $P(B) - P(A \cap B)$
- Neither (outside both): $1 - P(A \cup B)$
- Start with the central region $A \cap B \cap C$.
- Then fill the exclusive two-set intersections: $(A \cap B) - (A \cap B \cap C)$, etc.
- Then fill the regions belonging to exactly one set.
- Finally, calculate the "none of these" region.
For three sets $A$, $B$, $C$, the inclusion–exclusion principle gives:
$$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)$$Sample Space Diagrams
A sample space diagram (or two-way grid) systematically lists all possible outcomes of two combined experiments. Each cell represents one equally likely outcome. The probability of any event is the number of favourable cells divided by the total number of cells.
Conditional Probability
The conditional probability $P(A \mid B)$ is the probability that event $A$ occurs given that event $B$ is known to have occurred. Conditioning on $B$ restricts the sample space: we now only consider outcomes that lie within $B$, and ask what fraction of those also lie within $A$.
Frequency Trees
A frequency tree splits a population sequentially through a series of categories. Each branch shows frequencies (counts of people or items), not probabilities directly. Probabilities are then calculated as $\text{frequency on branch} \div \text{total}$. Frequency trees are especially useful for two-stage problems such as medical screening, quality control, or sequential selections.
🗺️ Visual Notes
- Scale: 0 = impossible, 1 = certain
- $P(A) + P(A') = 1$ (complement)
- Theoretical: equally likely outcomes
- Experimental: relative frequency from trials
- AND: $P(A \cap B) = P(A)P(B)$ — independent
- OR: $P(A \cup B) = P(A)+P(B)-P(A\cap B)$
- Mutually exclusive: $P(A\cap B)=0$
- Exhaustive: probabilities sum to 1
- 2-set: 4 regions including "neither"
- 3-set: 8 regions — fill from centre out
- Inclusion–exclusion for 3 sets
- Rectangle = entire sample space
- $P(A|B) = P(A \cap B) / P(B)$
- Restricts sample space to $B$
- Test independence: $P(A|B) = P(A)$?
- Without replacement creates dependence
- Sample space diagram (grid — two events)
- Tree diagram (sequential events)
- Frequency tree (populations)
- Two-way tables (cross-tabulation)
Two-Set Venn Diagram — Regions
Three-Set Venn Diagram — Worked Example (100 Students)
French ($F$): 60, Spanish ($S$): 50, German ($G$): 30, $F\cap S$: 25, $F\cap G$: 20, $S\cap G$: 15, all three: 10, neither: 10.
Key Probability Concepts Compared
| Concept | Definition | Key Formula / Condition | Example |
|---|---|---|---|
| Mutually Exclusive | Cannot both occur simultaneously | $P(A \cap B) = 0$ | Rolling a 3 and rolling a 5 on one die |
| Independent | One does not affect the other | $P(A \cap B) = P(A)P(B)$ | Tossing a coin and rolling a die |
| Exhaustive | Together they cover all outcomes | $\sum P(A_i) = 1$ | Getting H or T on a fair coin flip |
| Complementary | One is the negation of the other | $P(A') = 1 - P(A)$ | Raining / not raining tomorrow |
| Conditional | Probability given event $B$ occurred | $P(A|B) = P(A \cap B)/P(B)$ | P(2nd red | 1st was red), no replacement |
Which Rule to Use? — Decision Process
Sample Space Diagram — Two Dice (Sums)
The table shows the sum of two fair dice. There are 36 equally likely outcomes. Highlighted cells show sums of 7 — the most probable single sum.
| Die 1 \ Die 2 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| 6 | 7 | 8 | 9 | 10 | 11 | 12 |
$P(\text{sum} = 7) = \dfrac{6}{36} = \dfrac{1}{6}$ — the highest probability of any single sum.
✏️ Worked Examples
(a) $P(\text{red})$ (b) $P(\text{not blue})$ (c) $P(\text{red or green})$
(a) Find $P(A \cup B)$. (b) Determine whether $A$ and $B$ are independent, showing your working. (c) Find $P(A \mid B)$.
Calculate: $P(A) \times P(B) = 0.6 \times 0.5 = 0.30$
But $P(A \cap B) = 0.20 \neq 0.30$
Therefore $A$ and $B$ are not independent.
(a) Complete a 3-set Venn diagram showing the number of students in each region.
(b) Find $P(\text{student studies exactly one language})$.
(c) Find $P(F \mid S)$ and hence determine whether $F$ and $S$ are independent.
$F \cap S$ only (not $G$): $25 - 10 = 15$
$F \cap G$ only (not $S$): $20 - 10 = 10$
$S \cap G$ only (not $F$): $15 - 10 = 5$
Only $F$: $60 - 15 - 10 - 10 = 25$
Only $S$: $50 - 15 - 5 - 10 = 20$
Only $G$: $30 - 10 - 5 - 10 = 5$
Check total in diagram: $25 + 20 + 5 + 15 + 10 + 5 + 10 = 90$
Neither: $100 - 90 = 10$
Since $P(F \mid S) = 0.5 \neq 0.6 = P(F)$, knowing a student studies Spanish changes the probability of also studying French.
Verification via multiplication: $P(F) \times P(S) = 0.6 \times 0.5 = 0.30$, but $P(F \cap S) = 0.25 \neq 0.30$. ✓
Therefore $F$ and $S$ are not independent.
(b) $P(\text{exactly one}) = 0.5$
(c) $P(F \mid S) = 0.5$; $F$ and $S$ are not independent since $P(F \mid S) = 0.5 \neq 0.6 = P(F)$, equivalently $P(F \cap S) = 0.25 \neq 0.30 = P(F)P(S)$
❓ Exam Questions
The probability that a bus arrives late is 0.08. What is the probability that the bus does not arrive late?
Mark scheme: B1 for 0.92
A spinner has 5 equal sections numbered 1 to 5. A fair coin is tossed and the spinner is spun independently. Find the probability of obtaining a head and an even number.
The coin toss and spinner are independent events.
$P(\text{head}) = \dfrac{1}{2}$ $P(\text{even}) = \dfrac{2}{5}$ (the even numbers are 2 and 4)
$$P(\text{head} \cap \text{even}) = \frac{1}{2} \times \frac{2}{5} = \frac{2}{10} = \frac{1}{5}$$ Mark scheme: M1 for $\dfrac{1}{2} \times \dfrac{2}{5}$ (or equivalent); A1 for $\dfrac{1}{5}$ (oe)
$P(A) = 0.7$, $P(B) = 0.4$, $P(A \cap B) = 0.3$.
(a) Find $P(A \cup B)$. (b) State, with a reason, whether $A$ and $B$ are mutually exclusive.
(b) $A$ and $B$ are not mutually exclusive because $P(A \cap B) = 0.3 \neq 0$. Mutually exclusive events cannot occur simultaneously, which requires $P(A \cap B) = 0$.
Mark scheme: M1 for correct substitution in OR rule; A1 for 0.8; B1 for correct statement with reason
In a class of 80 students, 35 own a cat ($C$), 28 own a dog ($D$), and 12 own both a cat and a dog.
(a) Complete a Venn diagram to represent this information.
(b) Find $P(C \mid D)$.
Only $C$: $35 - 12 = 23$ $C \cap D$: $12$ Only $D$: $28 - 12 = 16$ Neither: $80 - 23 - 12 - 16 = 29$
Check: $23 + 12 + 16 + 29 = 80$ ✓
(b) $$P(C \mid D) = \frac{P(C \cap D)}{P(D)} = \frac{12/80}{28/80} = \frac{12}{28} = \frac{3}{7}$$ Mark scheme: B1 for each correct region (max B2 for Venn); M1 for $\dfrac{12}{28}$ or equivalent; A1 for $\dfrac{3}{7}$
A bag contains 4 red and 6 blue counters. Two counters are taken from the bag without replacement.
(a) Find $P(\text{both red})$. (b) Find $P(\text{exactly one red})$.
(c) Find $P(\text{second is blue} \mid \text{first was red})$.
(d) Determine, with a full calculation, whether "first counter is red" and "second counter is red" are independent events.
$P(A) = \dfrac{4}{10} = \dfrac{2}{5}$
$P(B) = P(RR) + P(BR) = \dfrac{12}{90} + \dfrac{24}{90} = \dfrac{36}{90} = \dfrac{2}{5}$
$P(A) \times P(B) = \dfrac{2}{5} \times \dfrac{2}{5} = \dfrac{4}{25} = 0.16$
$P(A \cap B) = \dfrac{12}{90} = \dfrac{2}{15} \approx 0.133$
Since $\dfrac{2}{15} \neq \dfrac{4}{25}$, the events are not independent. Drawing without replacement creates dependence.
Mark scheme: M1A1 for (a); M1A1 for (b); B1 for (c); M1 for comparing $P(A \cap B)$ with $P(A)P(B)$; A1 for correct conclusion with supporting calculation
A biased coin has probability $p$ of landing heads. The coin is flipped twice, independently. Given that $P(\text{exactly one head}) = 0.42$, find all possible values of $p$.
$P(\text{exactly one head}) = P(HT) + P(TH) = p(1-p) + (1-p)p = 2p(1-p)$
Setting equal to 0.42: $$2p(1-p) = 0.42 \implies 2p - 2p^2 = 0.42 \implies p^2 - p + 0.21 = 0$$ Using the quadratic formula (or factorising): $(p - 0.3)(p - 0.7) = 0$ $$p = 0.3 \quad \text{or} \quad p = 0.7$$ Both values are valid probabilities. The two solutions correspond to the same physical situation by symmetry (a coin with $P(H) = 0.3$ is the "complement" of one with $P(H) = 0.7$).
Mark scheme: M1 for setting $2p(1-p) = 0.42$; M1 for correct quadratic and attempt to solve; A1 for both $p = 0.3$ and $p = 0.7$
⭐ Grade 9 Model Answers
The following is a full annotated model answer for Q5 — the 6-mark question on without-replacement probability and independence. This type of question distinguishes Grade 8 from Grade 9 performance.
Part (a) — $P(\text{both red})$:
The critical insight is recognising that drawing without replacement means the second draw is dependent on the first. After removing one red counter, the bag contains 9 counters (3 red, 6 blue).
$$P(RR) = P(\text{1st red}) \times P(\text{2nd red} \mid \text{1st red}) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$$Mark-earning move: Writing both fractions explicitly and multiplying. Stating "$\frac{3}{9}$ because one red has been removed" shows clear understanding.
Part (b) — $P(\text{exactly one red})$:
Exactly one red means either Red-then-Blue or Blue-then-Red. These outcomes are mutually exclusive, so their probabilities add.
$$P(RB) = \frac{4}{10} \times \frac{6}{9} = \frac{24}{90}, \qquad P(BR) = \frac{6}{10} \times \frac{4}{9} = \frac{24}{90}$$ $$P(\text{exactly one red}) = \frac{24}{90} + \frac{24}{90} = \frac{48}{90} = \frac{8}{15}$$Mark-earning move: Considering both orderings (RB and BR). A common error is to write only $P(RB)$ and double it without recognising these are two distinct events.
Part (c) — Conditional probability:
Conditioning on "first was red" immediately tells us the new state of the bag: 9 counters remain with 3 red and 6 blue. No formula is needed — just read from the restricted sample space.
$$P(\text{2nd blue} \mid \text{1st red}) = \frac{6}{9} = \frac{2}{3}$$Mark-earning move: Reducing both numerator and denominator after the first draw. Writing $\frac{6}{10}$ (ignoring the condition) loses the mark.
Part (d) — Testing independence (Grade 9 discriminating skill):
Independence requires $P(A \cap B) = P(A) \times P(B)$ where $A$ = "1st is red" and $B$ = "2nd is red". A common error at Grade 8 is to state "not independent because it is without replacement" without calculation — this alone does not earn full marks.
$P(A) = \dfrac{4}{10} = \dfrac{2}{5}$
$P(B) = P(\text{2nd red}) = P(RR) + P(BR) = \dfrac{12}{90} + \dfrac{24}{90} = \dfrac{36}{90} = \dfrac{2}{5}$
$P(A) \times P(B) = \dfrac{2}{5} \times \dfrac{2}{5} = \dfrac{4}{25}$
$P(A \cap B) = P(RR) = \dfrac{2}{15}$
Convert to a common denominator to compare: $\dfrac{4}{25} = \dfrac{12}{75}$ and $\dfrac{2}{15} = \dfrac{10}{75}$. Since $\dfrac{12}{75} \neq \dfrac{10}{75}$, the events are not independent.
Mark-earning moves: (i) Correctly computing $P(\text{2nd red})$ by combining both routes (RR and BR); (ii) Computing both $P(A)P(B)$ and $P(A \cap B)$ numerically; (iii) Stating a clear conclusion. This full working earns both the method mark and the accuracy mark.
📋 Revision Sheet
| Term | Meaning |
|---|---|
| Event | A specific outcome or set of outcomes |
| Sample Space | The set of all possible outcomes |
| Mutually Exclusive | Cannot both occur; $P(A \cap B) = 0$ |
| Exhaustive | Cover all possibilities; sum of probs = 1 |
| Independent | One does not affect the other |
| Dependent | One affects the probability of the other |
| Relative Frequency | Experimental estimate of probability |
| Conditional Prob. | $P(A|B)$: prob. of $A$ given $B$ occurred |
$P(A') = 1 - P(A)$
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$P(A \cap B) = P(A) \times P(B)$ (independent only)
$P(A \cap B) = P(A) \times P(B \mid A)$ (general)
$P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$
Independence: $P(A \cap B) = P(A)P(B)$?
3-set: $P(A \cup B \cup C) = P(A)+P(B)+P(C)$
$\quad -P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)$
- AND → multiply (for independent events)
- OR → add, then subtract overlap
- NOT → 1 minus (complement rule)
- "Given B" → zoom into B's circle
- Exclusive → no overlap (can't both happen)
- 3-set Venn → fill from the centre out
- Without replacement → always dependent
- More trials → relative freq. → true prob.
- Always verify independence before using $P(A)P(B)$
- "Without replacement" signals dependent events — update the sample space at each stage
- In Venn diagram questions, check all regions sum to the total population
- For 3-set Venn diagrams, start with the central region $A \cap B \cap C$
- To test independence: compute $P(A)P(B)$ and compare to $P(A \cap B)$ numerically
- State conclusions explicitly: "Therefore $A$ and $B$ are / are not independent because..."
- "Given that" always signals $P(A \mid B)$ — the conditioning event goes in the denominator
- Mutually exclusive and independent are different concepts — never confuse them
🔄 Flashcards
Click a card to reveal the answer. Review all 15 cards regularly — spaced repetition is the most effective way to retain probability formulae.
✗ Common Mistakes
What students do: Write $P(A \cup B) = P(A) + P(B)$ without subtracting $P(A \cap B)$.
Why marks are lost: This double-counts every outcome in $A \cap B$, giving a result that is too large. If $P(A \cup B) > 1$, the error is immediately obvious — but when the answer looks plausible, students miss it entirely.
How to avoid it: Memorise the full OR rule: "add both, subtract the middle". Only skip the subtraction after explicitly confirming $P(A \cap B) = 0$ (mutually exclusive events).
What students do: Write $P(A \cap B) = P(A) \times P(B)$ even when drawing without replacement.
Why marks are lost: The formula only holds for independent events. For dependent events, the correct formula is $P(A \cap B) = P(A) \times P(B \mid A)$, and $P(B \mid A) \neq P(B)$ once items have been removed from the pool.
How to avoid it: Ask "does the first event change what is available for the second?" If yes (e.g., "without replacement", "selected and not returned"), use conditional probabilities on each branch of a tree diagram.
What students do: Place "25 study $F$ and $S$" entirely in the $F \cap S$ crescent region, without first removing those who study all three.
Why marks are lost: The given value for $F \cap S$ includes the triple intersection. The exclusive two-set overlap is $(F \cap S) - (F \cap S \cap G) = 25 - 10 = 15$. Students who skip this end up with inflated counts that do not sum to the total.
How to avoid it: Always fill from the centre outwards. The centre ($A \cap B \cap C$) value is subtracted from each two-set intersection to get the exclusive crescent.
What students do: Swap the numerator and denominator, or treat $P(A \mid B)$ and $P(B \mid A)$ as equal.
Why marks are lost: These are generally very different values. For example, $P(\text{disease} \mid \text{positive test})$ and $P(\text{positive test} \mid \text{disease})$ are completely different quantities with entirely different real-world meanings.
How to avoid it: In $P(A \mid B) = P(A \cap B) / P(B)$, the event after the vertical bar is always the denominator. Write the formula out fully before substituting.
What students do: Write "the events are mutually exclusive, so they are independent."
Why marks are lost: This is the opposite of the truth. If $A$ and $B$ are mutually exclusive with $P(A) > 0$ and $P(B) > 0$, knowing $A$ occurred guarantees $B$ did not — the events are as dependent as possible. Independence and mutual exclusivity are fundamentally different properties.
How to avoid it: Mutually exclusive: $P(A \cap B) = 0$. Independent: $P(A \cap B) = P(A)P(B)$. Both can only hold simultaneously if $P(A) = 0$ or $P(B) = 0$.
What students do: Calculate only $P(RB)$ when asked for $P(\text{exactly one red})$, forgetting $P(BR)$.
Why marks are lost: "Exactly one red" can happen two ways: Red first then Blue, or Blue first then Red. Only one of these orderings is considered, halving the correct answer. Examiners set these questions precisely to test this point.
How to avoid it: Always draw a tree diagram and identify all branches that satisfy the condition. Add all such branches (they are mutually exclusive outcomes of the combined experiment).
✅ Final Checklist
Click each item when you are confident with it. Your progress is saved automatically in your browser.
- I can place any event on the probability scale from 0 (impossible) to 1 (certain)
- I can calculate theoretical probability using the formula: favourable outcomes ÷ total equally likely outcomes
- I can calculate relative frequency from experimental data and explain why it estimates theoretical probability
- I understand the Law of Large Numbers: relative frequency converges to true probability as trials increase
- I can apply the complement rule $P(A') = 1 - P(A)$ to find "not A" probabilities
- I can identify mutually exclusive events and apply the simplified OR rule $P(A \cup B) = P(A) + P(B)$
- I can apply the full OR rule $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ for any two events
- I can apply the AND rule $P(A \cap B) = P(A) \times P(B)$ for independent events
- I can explain why "without replacement" makes events dependent and use conditional probabilities on tree diagrams
- I can draw and interpret a 2-set Venn diagram, correctly calculating all four regions
- I can complete a 3-set Venn diagram by filling from the centre outwards and verify all regions sum to the total
- I can state and apply the inclusion–exclusion formula for three sets
- I can calculate conditional probability using $P(A \mid B) = P(A \cap B) / P(B)$ from a Venn diagram or given probabilities
- I can test whether two events are independent by checking $P(A \cap B) = P(A)P(B)$ and state a clear conclusion
- I can solve multi-step Grade 9 probability problems combining Venn diagrams, conditional probability, and independence testing