What is the basic definition of probability in Statistics 1 for Edexcel International A Levels?

In the context of Statistics 1 for Edexcel International A Levels, probability is defined as a measure of the likelihood that a particular event will occur. It is quantified as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. Probability can be calculated using the formula: Probability of an event = Number of favorable outcomes / Total number of possible outcomes.

How do you calculate the probability of multiple independent events occurring?

To calculate the probability of multiple independent events occurring, you multiply the probabilities of each individual event. For example, if you want to find the probability of two independent events A and B both occurring, you would use the formula: P(A and B) = P(A) * P(B). This principle is crucial in the Edexcel International A Levels Statistics 1 curriculum when dealing with compound events.

What is the difference between mutually exclusive and independent events in probability?

Mutually exclusive events are events that cannot occur at the same time. For example, when flipping a coin, the events 'heads' and 'tails' are mutually exclusive. Independent events, on the other hand, are events where the occurrence of one event does not affect the probability of the other. Understanding these concepts is essential in the Edexcel International A Levels Statistics 1 syllabus, as they form the basis for more complex probability calculations.

How is conditional probability different from regular probability?

Conditional probability refers to the probability of an event occurring given that another event has already occurred. It is calculated using the formula: P(A|B) = P(A and B) / P(B), where P(A|B) is the probability of event A occurring given that B has occurred. This concept is a key part of the Edexcel International A Levels Statistics 1 curriculum, as it helps in understanding the relationships between dependent events.

What role do probability distributions play in Statistics 1 for Edexcel International A Levels?

Probability distributions describe how probabilities are distributed over the values of a random variable. They are fundamental in Statistics 1 for Edexcel International A Levels, as they provide a framework for understanding how different outcomes are likely to occur. Common probability distributions include the binomial distribution and the normal distribution, each with specific properties and applications in statistical analysis.

Why is "Confusing mutually exclusive with independent" a common mistake in Probability?

Why it happens: Both phrases sound like 'unrelated'. How to avoid it: Mutually exclusive: P(A B) = 0 (they cannot both happen). Independent: P(A B) = P(A) · P(B) (they don't influence each other). If both events have positive probability and they are mutually exclusive, they are NOT independent. Source: WST01/01 examiner reports — annual

Why is "Using $P(A) + P(B)$ for $P(A \cap B)$" a common mistake in Probability?

Why it happens: Confusion between AND and OR. How to avoid it: OR (union): addition rule with subtraction of overlap. AND (intersection): multiplication if independent, or use conditional formula otherwise.

Why is "Computing $P(A) \cdot P(B)$ for $P(A \cap B)$ when not independent" a common mistake in Probability?

Why it happens: Default to multiplication. How to avoid it: Check independence first. If independent, multiply. If not, use the addition rule or conditional formula: P(A B) = P(A) P(B|A) = P(B) P(A|B).

Why is "Computing $P(D|X)$ when $P(X|D)$ is asked" a common mistake in Probability?

Why it happens: Conditional direction reversed. How to avoid it: Read 'given that' carefully: 'given the component is defective, what is the probability it came from X' is P(X|D). Use P(X|D) = P(X D)/P(D). Source: WST01/01 examiner reports — recurring

Why is "Starting Venn-diagram regions from the outside" a common mistake in Probability?

Why it happens: Filling in the easiest region first. How to avoid it: Always start with the triple intersection (the centre), then pair intersections, then single events, then outside. Subtract along the way.

Why is "Treating branches at the same level as 'same probability'" a common mistake in Probability?

Why it happens: Visual symmetry of trees masks conditioning. How to avoid it: Each branch is conditional on the path taken to reach it. The two branches from a single node sum to 1; branches at the same depth do NOT.

Statistics 1Edexcel International A Levels Mathematics (XMA01-YMA01)

Probability

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Probability Study Notes — Edexcel IAL Mathematics S1 (WST01/01, 2018 spec — 2026 onwards)

Sample spaces, events, set notation, addition and multiplication rules, mutual exclusion, independence, conditional probability, Venn and tree diagrams, the law of total probability. The grammar of every later S1 chapter.

What you’ll learn

Mapped to the Cambridge IGCSE XMA01-YMA01 syllabus (2026 onwards).

S1 4.1 — Understand sample spaces, events, and set notation including $\cup$ , $\cap$ and $'$ .
S1 4.2 — Use the addition rule for probabilities and identify mutually exclusive events.
S1 4.3 — Use conditional probability and identify independent events.
S1 4.4 — Use Venn diagrams to represent events and probabilities.
S1 4.5 — Use tree diagrams and the law of total probability for sequential events.

Sample spaces, events and set notation

Outcomes live in a sample space $S$ ; events are subsets; set notation describes them.

Definitions.

Sample space $S$ : set of all possible outcomes.
Event $A \subseteq S$ : a subset of interest.
Probability $P(A) \in [0, 1]$ with $P(S) = 1$ , $P(\emptyset) = 0$ .

Set notation.

Symbol	Meaning	Plain words
$A \cup B$	Union	$A$ OR $B$ (or both)
$A \cap B$	Intersection	$A$ AND $B$
$A'$	Complement	NOT $A$
$\emptyset$	Empty set	Impossible event
$A \subseteq B$	Subset	If $A$ happens then $B$ happens

Useful identities.

$P(A') = 1 - P(A)$ .
De Morgan: $(A \cup B)' = A' \cap B'$ , $(A \cap B)' = A' \cup B'$ .
$A = (A \cap B) \cup (A \cap B')$ — disjoint decomposition, hence $P(A) = P(A \cap B) + P(A \cap B')$ .

Equally likely outcomes. When the sample space is finite and outcomes are equally likely,

$P(A) = \dfrac{|A|}{|S|}.$

$0 \le P(A) \le 1$ ; $P(S) = 1$ ; $P(\emptyset) = 0$ .
$\cup$ = OR, $\cap$ = AND, $'$ = NOT.
$P(A') = 1 - P(A)$ — always.
De Morgan's laws connect complements of unions/intersections.

The addition rule

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$ — subtract the double-count.

Statement.

$P(A \cup B) = P(A) + P(B) - P(A \cap B).$

Why subtract? Adding $P(A) + P(B)$ counts the overlap $A \cap B$ twice. Subtracting it once corrects the double-count.

Mutually exclusive special case. If $A \cap B = \emptyset$ then $P(A \cap B) = 0$ and the rule simplifies to

$P(A \cup B) = P(A) + P(B).$

Worked example. $P(A) = 0.4$ , $P(B) = 0.5$ , $P(A \cap B) = 0.15$ .

$P(A \cup B) = 0.4 + 0.5 - 0.15 = 0.75$ .
$P(A' \cap B') = 1 - P(A \cup B) = 0.25$ (De Morgan).
$P(A \cap B') = P(A) - P(A \cap B) = 0.25$ .

Venn check: regions $A \cap B' = 0.25$ , $A \cap B = 0.15$ , $A' \cap B = 0.35$ , $A' \cap B' = 0.25$ . Total $= 1$ . ✓

Master formula: subtract the overlap once.
Mutually exclusive $\Rightarrow$ overlap is $0$ .
Complement: $P(A') = 1 - P(A)$ .
Disjoint decomposition: $P(A) = P(A \cap B) + P(A \cap B')$ .

Conditional probability and independence

$P(A|B)$ = restrict to the world where $B$ happened. Independent $\Leftrightarrow P(A|B) = P(A)$ .

Conditional probability.

$P(A | B) = \dfrac{P(A \cap B)}{P(B)}, \quad P(B) > 0.$

Interpretation: among the cases where $B$ has happened, what fraction also have $A$ ?

Multiplication form. Rearranging gives the chain rule:

$P(A \cap B) = P(B) \cdot P(A | B) = P(A) \cdot P(B | A).$

This is the formula behind every tree diagram (multiply along a branch path).

Independence. $A$ and $B$ are independent if either of the equivalent conditions holds:

$P(A \cap B) = P(A) P(B), \qquad P(A | B) = P(A), \qquad P(B | A) = P(B).$

Knowing that one event happened gives no information about the other.

Worked example. $P(A) = 0.6$ , $P(B) = 0.4$ , $P(A \cup B) = 0.8$ .

$P(A \cap B) = 0.6 + 0.4 - 0.8 = 0.2$ .
$P(A | B) = 0.2 / 0.4 = 0.5$ .
Test independence: $P(A) P(B) = 0.24 \neq 0.2$ — NOT independent.

Mutually exclusive vs independent — the key distinction. Two non-trivial events ( $P > 0$ ) cannot be both mutually exclusive AND independent, because mutual exclusion gives $P(A \cap B) = 0$ while independence gives $P(A \cap B) = P(A) P(B) > 0$ .

Conditional: $P(A|B) = P(A \cap B)/P(B)$ .
Chain rule: $P(A \cap B) = P(B) P(A|B) = P(A) P(B|A)$ .
Independent $\Leftrightarrow P(A \cap B) = P(A) P(B)$ (or $P(A|B) = P(A)$ ).
Mutually exclusive and independent are essentially contradictory for non-trivial events.

Venn diagrams — three-event case

Always fill from the centre outwards.

Procedure for three events $M, P, C$ .

Start with the triple intersection: $|M \cap P \cap C|$ .
Compute each pair-only region:
- $|M \cap P| - |M \cap P \cap C|$ etc.
Compute each single-event-only region:
- $|M| - (\text{the three regions inside } M\text{ that also touch } P \text{ or } C)$ .
Compute the outside: $|\text{Total}| - (\text{everything inside the three circles})$ .

Worked example. $60$ students; $30$ Maths, $25$ Physics, $20$ Chemistry; pair intersections $10, 8, 6$ ; all three $= 4$ .

Triple: $4$ .
Pair-only: $M \cap P$ only $= 6$ , $M \cap C$ only $= 4$ , $P \cap C$ only $= 2$ .
Single-only: $M = 30 - (6 + 4 + 4) = 16$ . $P = 25 - (6 + 4 + 2) = 13$ . $C = 20 - (4 + 4 + 2) = 10$ .
Outside: $60 - 55 = 5$ .

Useful regions.

'Exactly one of $M, P, C$ ' = sum of three single-only counts.
'Exactly two' = sum of three pair-only counts.
'All three' = triple intersection.
'At least one' = total inside circles.
'None' = outside.

Centre out: triple $\to$ pair-only $\to$ single-only $\to$ outside.
Probabilities: divide each region count by total.
Read 'exactly one/two/three' off the appropriate regions.
Sanity-check: all regions sum to total.

Tree diagrams and the law of total probability

Multiply along a branch; add over all branches; reverse with conditional formula.

Tree diagrams. Each node has children whose probabilities sum to $1$ . The probability of a particular sequence is the product of the conditional probabilities along the path.

Law of total probability. If $\{B_1, B_2, \ldots, B_k\}$ is a partition of the sample space (mutually exclusive and exhaustive), then for any event $D$ ,

$P(D) = \sum_{i = 1}^{k} P(B_i) \cdot P(D | B_i).$

This is the 'add over all branches that lead to $D$ ' rule.

Worked example (factory defects). Machines $X$ ( $60\%$ output, $3\%$ defect), $Y$ ( $40\%$ output, $5\%$ defect).

              0.03 D
        X ─────────┤
       /             0.97 D'
   0.6 │
       \           0.05 D
        Y ─────────┤
              0.95 D'

$P(X \cap D) = 0.6 \times 0.03 = 0.018$ .
$P(Y \cap D) = 0.4 \times 0.05 = 0.020$ .
$P(D) = 0.018 + 0.020 = 0.038$ .

Reversing the conditioning (Bayes-style). Given the defect, which machine made it?

$P(X | D) = \dfrac{P(X \cap D)}{P(D)} = \dfrac{0.018}{0.038} \approx 0.474.$

Even though $X$ produces more output, only about $47\%$ of defective items come from $X$ , because $Y$ 's defect rate is higher.

Branch probabilities at one node sum to $1$ .
Multiply along a path for joint probability.
Sum over all paths to $D$ for $P(D)$ (LTP).
Reverse direction with $P(B|D) = P(B \cap D)/P(D)$ .

How it’s examined

Probability is a fixture of WST01/01 — typically one or two questions worth $10$ - $14$ marks total per paper. Usual asks: addition-rule manipulations with given $P(A), P(B), P(A \cap B)$ ( $4$ - $5$ marks); conditional probability and an independence test ( $5$ - $6$ marks); three-event Venn diagram from a contextual scenario ( $6$ - $8$ marks); two-stage tree diagram with total probability and a reverse-Bayes sub-part ( $7$ marks); mutually-exclusive-vs-independent comparison ( $4$ - $5$ marks). Mark schemes routinely award B-marks for the explicit Venn or tree diagram itself.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WST01/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Step-by-step worked examples — Probability

Step-by-step solutions to past-paper-style questions on probability, written exactly the way a tutor would explain them at the board.

1Addition rule with Venn diagram (5 marks, S1)
Core• Adapted from WST01/01 January 2024 Q4• addition rule, Venn diagram
Question
$P(A) = 0.4$ , $P(B) = 0.5$ , $P(A \cap B) = 0.15$ . Find (a) $P(A \cup B)$ , (b) $P(A' \cap B')$ , (c) $P(A \cap B')$ .
Step-by-step solution
1. Step 1
  Apply the addition rule.
  $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.5 - 0.15 = 0.75$
2. Step 2
  By De Morgan's law, $A' \cap B' = (A \cup B)'$ .
  $P(A' \cap B') = 1 - P(A \cup B) = 1 - 0.75 = 0.25$
3. Step 3
  Decompose $A$ as the disjoint union $A = (A \cap B) \cup (A \cap B')$ .
  $P(A \cap B') = P(A) - P(A \cap B) = 0.4 - 0.15 = 0.25$
Answer
(a) $0.75$ . (b) $0.25$ . (c) $0.25$ .
Examiner tip
M1 for addition rule. A1 for $0.75$ . M1 for $1 - P(A \cup B)$ or equivalent. A1 for $0.25$ . M1 for $P(A) - P(A \cap B)$ . A1 for $0.25$ . Drawing the Venn diagram with the four region probabilities ( $0.25$ , $0.15$ , $0.35$ , $0.25$ ) summing to $1$ is a strong sanity check.
2Conditional probability and independence (6 marks, S1)
Extended• Adapted from WST01/01 June 2024 Q7• conditional, independence
Question
$P(A) = 0.6$ , $P(B) = 0.4$ , $P(A \cup B) = 0.8$ . (a) Find $P(A \cap B)$ . (b) Find $P(A|B)$ . (c) Determine, with reason, whether $A$ and $B$ are independent.
Step-by-step solution
1. Step 1
  Rearrange the addition rule:
  $P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.6 + 0.4 - 0.8 = 0.2$
2. Step 2
  Apply the conditional formula:
  $P(A|B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{0.2}{0.4} = 0.5$
3. Step 3
  Independence test: $A \perp B \Leftrightarrow P(A \cap B) = P(A) \cdot P(B)$ .
  $P(A) \cdot P(B) = 0.6 \times 0.4 = 0.24$
4. Step 4
  Since $P(A \cap B) = 0.2 \neq 0.24 = P(A) \cdot P(B)$ , the events are not independent. (Equivalently: $P(A|B) = 0.5 \neq 0.6 = P(A)$ .)
Answer
(a) $P(A \cap B) = 0.2$ . (b) $P(A|B) = 0.5$ . (c) Not independent since $P(A \cap B) \neq P(A) \cdot P(B)$ .
Examiner tip
M1 for addition-rule rearrangement. A1 for $0.2$ . M1 for conditional formula. A1 for $0.5$ . B1 for stating the independence test. A1 for the correct conclusion with reason. Mark schemes require both the calculation AND the verbal conclusion in (c).
3Tree diagram with conditional probabilities (7 marks, S1)
Extended• Adapted from WST01/01 January 2023 Q7• tree diagram, conditional, total probability
Question
A factory has two machines. Machine $X$ produces $60\%$ of the output, with $3\%$ defective. Machine $Y$ produces $40\%$ , with $5\%$ defective. A component is selected at random. (a) Find the probability it is defective. (b) Given that it is defective, find the probability it came from machine $X$ .
Step-by-step solution
1. Step 1
  Set up. Define events $X$ = made by $X$ , $Y$ = made by $Y$ , $D$ = defective. Given: $P(X) = 0.6$ , $P(Y) = 0.4$ , $P(D|X) = 0.03$ , $P(D|Y) = 0.05$ .
2. Step 2
  Law of total probability.
  $P(D) = P(X)P(D|X) + P(Y)P(D|Y) = 0.6 \times 0.03 + 0.4 \times 0.05$
3. Step 3
  Compute:
  $P(D) = 0.018 + 0.020 = 0.038$
4. Step 4
  Apply Bayes' (or conditional) formula:
  $P(X|D) = \dfrac{P(X \cap D)}{P(D)} = \dfrac{0.018}{0.038} = \dfrac{9}{19}$
5. Step 5
  Decimal:
  $P(X|D) \approx 0.474 \;(3\,\text{sf})$
Answer
(a) $P(D) = 0.038$ . (b) $P(X|D) = 9/19 \approx 0.474$ (3 sf).
Examiner tip
M1 for tree diagram or LTP setup. M1 for the products $0.6 \times 0.03$ and $0.4 \times 0.05$ . A1 for $P(D) = 0.038$ . M1 for the conditional formula. A1 for $P(X|D)$ as exact fraction. A1 for the decimal to $3$ sf. Drawing the tree (and labelling each branch with its probability) is a B1 mark in some mark schemes.
4Venn diagram with three sets (7 marks, S1)
Extended• Adapted from WST01/01 June 2023 Q8• Venn diagram, three events
Question
$60$ students take subjects from Maths $(M)$ , Physics $(P)$ and Chemistry $(C)$ . $30$ take Maths, $25$ Physics, $20$ Chemistry. $10$ take Maths and Physics, $8$ Maths and Chemistry, $6$ Physics and Chemistry. $4$ take all three. (a) Draw a Venn diagram showing the number in each region. (b) Find $P(\text{exactly two subjects})$ .
Step-by-step solution
1. Step 1
  Start at the centre: $|M \cap P \cap C| = 4$ .
2. Step 2
  Pair intersections: $|M \cap P| = 10 \Rightarrow$ region with $M \cap P$ only $= 10 - 4 = 6$ . $|M \cap C| - 4 = 4$ in the $M \cap C$ only region. $|P \cap C| - 4 = 2$ in the $P \cap C$ only region.
3. Step 3
  Single-subject regions. $M$ only $= 30 - (6 + 4 + 4) = 16$ . $P$ only $= 25 - (6 + 4 + 2) = 13$ . $C$ only $= 20 - (4 + 4 + 2) = 10$ .
4. Step 4
  Total inside the three circles: $16 + 13 + 10 + 6 + 4 + 2 + 4 = 55$ . So $60 - 55 = 5$ students are outside (take none of the three).
5. Step 5
  Exactly two subjects = sum of the three 'pair-only' regions: $6 + 4 + 2 = 12$ .
  $P(\text{exactly two}) = \dfrac{12}{60} = 0.2$
Answer
Regions (clockwise from $M$ -only): $16$ , $6$ , $13$ , $2$ , $10$ , $4$ , $4$ inside; $5$ outside. $P(\text{exactly two}) = 12/60 = 0.2$ .
Examiner tip
B1 for starting from the centre $= 4$ . M1 for pair-only regions. A1 for all three. M1 for single-subject regions. A1 for all three. B1 for the outside count $5$ . M1A1 for $P(\text{exactly two}) = 0.2$ . Working from the centre outwards is the only reliable method.
5Mutually exclusive vs independent (4 marks, S1)
Core• Adapted from WST01/01 June 2024 Q8• mutually exclusive, independent
Question
$A$ and $B$ are mutually exclusive events with $P(A) = 0.3$ , $P(B) = 0.5$ . (a) Find $P(A \cup B)$ . (b) Determine whether $A$ and $B$ can be independent. (c) Find $P(A | B)$ .
Step-by-step solution
1. Step 1
  Mutually exclusive: $P(A \cap B) = 0$ , so the addition rule simplifies.
  $P(A \cup B) = P(A) + P(B) = 0.3 + 0.5 = 0.8$
2. Step 2
  If $A$ and $B$ were independent, $P(A \cap B) = P(A) \cdot P(B) = 0.15 \neq 0$ . But mutually exclusive forces $P(A \cap B) = 0$ . So $A$ and $B$ cannot be both mutually exclusive AND independent (unless one of them has probability $0$ ).
3. Step 3
  Conditional:
  $P(A | B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{0}{0.5} = 0$
4. Step 4
  Interpretation: if $B$ has happened, then $A$ cannot have happened too (because they exclude each other), so $P(A|B) = 0$ .
Answer
(a) $0.8$ . (b) Not independent (mutual exclusion gives $P(A \cap B) = 0$ , but independence would require $0.15$ ). (c) $0$ .
Examiner tip
B1 for $P(A \cap B) = 0$ . M1A1 for $0.8$ . B1 for the explicit comparison $0.15$ vs $0$ . A1 for the conclusion. B1 for $P(A|B) = 0$ . Common student error: assuming mutually exclusive $\Rightarrow$ independent. Examiners flag this annually.

Key Formulae — Probability

The formulae you need to memorise for probability on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Addition rule
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$A, B$
events
$\cup$
union (or)
$\cap$
intersection (and)
When to use
Whenever you need $P(A \cup B)$ , $P(A \cap B)$ or $P(A \cap B')$ given the others. The 'master formula' for two-event problems. IS in the IAL formula booklet.
Example
$P(A) = 0.4$ , $P(B) = 0.5$ , $P(A \cap B) = 0.15$ : $P(A \cup B) = 0.75$ .
Conditional probability
$P(A | B) = \dfrac{P(A \cap B)}{P(B)}, \quad P(B) > 0$
$P(A|B)$
probability of $A$ given $B$
When to use
Reverse conditioning (Bayes), tree-diagram analyses, sequential experiments where the outcome of one stage influences the next. IS in the IAL formula booklet.
Example
$P(A \cap B) = 0.2$ , $P(B) = 0.4$ : $P(A|B) = 0.5$ .
Independence
$A, B \text{ independent} \Leftrightarrow P(A \cap B) = P(A) \cdot P(B)$
$A \perp B$
$A$ is independent of $B$
When to use
Testing whether two events are independent. Equivalent statements: $P(A|B) = P(A)$ , $P(B|A) = P(B)$ . NOT in the booklet — definitional.
Example
$P(A) = 0.6$ , $P(B) = 0.4$ , $P(A \cap B) = 0.24 = 0.6 \times 0.4$ : independent.
Law of total probability
$P(D) = P(X) P(D|X) + P(Y) P(D|Y) + \ldots$
$X, Y, \ldots$
mutually exclusive, exhaustive events
$D$
the event of interest
When to use
When the probability of $D$ depends on which 'source' or 'route' is used to reach it. Standard for tree diagrams. NOT in the booklet under this name but is given as a special case in the booklet's conditional / multiplication formulas.
Example
Defect probabilities by machine: $P(D) = 0.6 \times 0.03 + 0.4 \times 0.05 = 0.038$ .

Key Definitions and Keywords — Probability

Definitions to memorise and the exact keywords mark schemes credit for probability answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Sample space
Examiner keyword
The set $S$ of all possible outcomes of an experiment. Probability is a function on subsets (events) of $S$ that assigns each a number in $[0, 1]$ with $P(S) = 1$ .
Event
Examiner keyword
A subset of the sample space. 'Rolling an even number on a die' is the event $\{2, 4, 6\}$ .
Union, intersection, complement
Examiner keyword
$A \cup B$ = all outcomes in $A$ or $B$ (or both). $A \cap B$ = outcomes in both $A$ and $B$ . $A'$ = outcomes NOT in $A$ . Together they describe any region of a Venn diagram.
Mutually exclusive
Examiner keyword
Events that cannot both happen: $P(A \cap B) = 0$ . Equivalently, $A$ and $B$ are disjoint subsets of the sample space.
Independent
Examiner keyword
Events for which the occurrence of one does not affect the probability of the other: $P(A \cap B) = P(A) \cdot P(B)$ , equivalently $P(A|B) = P(A)$ . Distinct from mutually exclusive — and the two cannot generally coexist.
Conditional probability
Examiner keyword
$P(A | B)$ = probability that $A$ happens given that $B$ has happened. Defined as $P(A \cap B) / P(B)$ .
Tree diagram
Branching diagram showing sequential events. Each branch is labelled with a conditional probability; the probability of a sequence is the product along the branch path.

Common Mistakes and Misconceptions — Probability

The traps other students keep falling into on probability questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Confusing mutually exclusive with independent
WST01/01 examiner reports — annual
Why it happens
Both phrases sound like 'unrelated'.
How to avoid it
Mutually exclusive: $P(A \cap B) = 0$ (they cannot both happen). Independent: $P(A \cap B) = P(A) \cdot P(B)$ (they don't influence each other). If both events have positive probability and they are mutually exclusive, they are NOT independent.
✕Using $P(A) + P(B)$ for $P(A \cap B)$
Why it happens
Confusion between AND and OR.
How to avoid it
OR (union): addition rule with subtraction of overlap. AND (intersection): multiplication if independent, or use conditional formula otherwise.
✕Computing $P(A) \cdot P(B)$ for $P(A \cap B)$ when not independent
Why it happens
Default to multiplication.
How to avoid it
Check independence first. If independent, multiply. If not, use the addition rule or conditional formula: $P(A \cap B) = P(A) P(B|A) = P(B) P(A|B)$ .
✕Computing $P(D|X)$ when $P(X|D)$ is asked
WST01/01 examiner reports — recurring
Why it happens
Conditional direction reversed.
How to avoid it
Read 'given that' carefully: 'given the component is defective, what is the probability it came from $X$ ' is $P(X|D)$ . Use $P(X|D) = P(X \cap D)/P(D)$ .
✕Starting Venn-diagram regions from the outside
Why it happens
Filling in the easiest region first.
How to avoid it
Always start with the triple intersection (the centre), then pair intersections, then single events, then outside. Subtract along the way.
✕Treating branches at the same level as 'same probability'
Why it happens
Visual symmetry of trees masks conditioning.
How to avoid it
Each branch is conditional on the path taken to reach it. The two branches from a single node sum to $1$ ; branches at the same depth do NOT.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Video lesson

Short walkthrough of the concepts students most often get stuck on.

Probability — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Probability

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Addition rule with Venn diagram (5 marks, S1)

2Conditional probability and independence (6 marks, S1)

3Tree diagram with conditional probabilities (7 marks, S1)

4Venn diagram with three sets (7 marks, S1)

5Mutually exclusive vs independent (4 marks, S1)

Addition rule

Conditional probability

Independence

Law of total probability

Sample space

Event

Union, intersection, complement

Mutually exclusive

Independent

Conditional probability

Tree diagram

✕Confusing mutually exclusive with independent

✕Using P(A)+P(B)P(A) + P(B)P(A)+P(B) for P(A∩B)P(A \cap B)P(A∩B)

✕Computing P(A)⋅P(B)P(A) \cdot P(B)P(A)⋅P(B) for P(A∩B)P(A \cap B)P(A∩B) when not independent

✕Computing P(D∣X)P(D|X)P(D∣X) when P(X∣D)P(X|D)P(X∣D) is asked

✕Starting Venn-diagram regions from the outside

✕Treating branches at the same level as 'same probability'

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Probability — frequently asked questions

✕Using $P(A) + P(B)$ for $P(A \cap B)$

✕Computing $P(A) \cdot P(B)$ for $P(A \cap B)$ when not independent

✕Computing $P(D|X)$ when $P(X|D)$ is asked