Probability

What you'll learn

Probability appears in both Paper 2 and Paper 4 of CIE IGCSE Mathematics, typically accounting for 8-12 marks per paper. This topic requires you to calculate the likelihood of events occurring, analyse combined events using sample spaces, tree diagrams and Venn diagrams, and interpret probability in real-world contexts. Questions frequently combine probability with algebra, fractions, and set notation.

Key terms and definitions

Probability — a numerical measure between 0 and 1 (or 0% to 100%) representing how likely an event is to occur, calculated as favourable outcomes divided by total possible outcomes.

Event — a specific outcome or combination of outcomes from a probability experiment, such as rolling a 6 on a die or drawing a red card.

Mutually exclusive events — two or more events that cannot occur simultaneously; if one happens, the others cannot (e.g., rolling a 3 or a 5 on a single die roll).

Independent events — two events where the outcome of one does not affect the probability of the other (e.g., tossing a coin twice; the second toss is independent of the first).

Complementary events — two mutually exclusive events that together cover all possible outcomes; if P(A) is the probability of event A, then P(A') = 1 - P(A).

Sample space — the complete set of all possible outcomes for a probability experiment, often displayed in a list, table, or diagram.

Expected frequency — the theoretical number of times an outcome should occur in a given number of trials, calculated as probability × number of trials.

Conditional probability — the probability of an event occurring given that another event has already occurred, written as P(A|B) and relevant when events are not independent.

Core concepts

Basic probability calculations

The fundamental probability formula appears in virtually every CIE IGCSE probability question:

P(event) = Number of favourable outcomes / Total number of possible outcomes

Probability values always satisfy: 0 ≤ P(event) ≤ 1

• P(event) = 0 means the event is impossible
• P(event) = 1 means the event is certain
• P(event) = 0.5 means the event has an equal chance of occurring or not occurring

Probabilities can be expressed as fractions, decimals, or percentages. CIE mark schemes typically accept any of these forms unless the question specifies otherwise.

For complementary events: P(A) + P(A') = 1, which rearranges to P(A') = 1 - P(A). This relationship is particularly useful when calculating "at least one" probabilities or when finding P(not A) is simpler than finding P(A) directly.

Expected frequency and relative frequency

Expected frequency connects theoretical probability to practical outcomes:

Expected frequency = Probability × Number of trials

For example, if P(rolling a 6) = 1/6 and you roll a die 120 times, the expected frequency = 1/6 × 120 = 20.

Relative frequency is experimental probability based on actual results:

Relative frequency = Number of times event occurred / Total number of trials

CIE questions often ask you to compare theoretical probability with relative frequency from experiments, or use large-sample relative frequency to estimate probability when theoretical calculation is impossible.

Combined events and sample spaces

When two or more events occur together, representing all possible outcomes systematically prevents errors.

Two-way tables work well for two events with few outcomes:

	Heads	Tails
1	(1,H)	(1,T)
2	(2,H)	(2,T)
3	(3,H)	(3,T)
4	(4,H)	(4,T)
5	(5,H)	(5,T)
6	(6,H)	(6,T)

List notation can show sample spaces concisely: S = {(1,H), (1,T), (2,H), (2,T), ..., (6,T)}.

For mutually exclusive events that cover all outcomes:

P(A or B) = P(A) + P(B)

For example, the probability of rolling a 2 or a 5 on a fair die = 1/6 + 1/6 = 2/6 = 1/3.

For independent events occurring together:

P(A and B) = P(A) × P(B)

For example, the probability of flipping heads twice = 1/2 × 1/2 = 1/4.

Tree diagrams

Tree diagrams are essential for CIE IGCSE probability questions involving sequential events or conditional probability. They systematically display all possible outcome paths.

Constructing tree diagrams:

1. Draw branches for the first event, labelling each outcome and writing its probability on the branch
2. From each first-event outcome, draw branches for the second event with appropriate probabilities
3. Continue for additional events if needed
4. Multiply probabilities along each complete path to find P(that sequence)
5. Add probabilities of different paths that satisfy the required condition

With replacement (independent events): probabilities remain constant on second and subsequent branches.

Without replacement (dependent events): probabilities change on subsequent branches. If a bag contains 5 red and 3 blue balls, P(first red) = 5/8, but P(second red | first red) = 4/7 because one red ball has been removed.

CIE questions commonly require tree diagrams for:

• Drawing objects from bags or boxes without replacement
• Two-stage games or tests
• Quality control scenarios with defective items
• Weather patterns or journey routes with conditional probabilities

Venn diagrams and set notation

Venn diagrams appear frequently in CIE IGCSE Mathematics Paper 4, combining probability with set theory.

Set notation in probability:

• P(A ∪ B) = probability of A or B or both occurring
• P(A ∩ B) = probability of both A and B occurring
• P(A') = probability of A not occurring
• n(A) = number of elements in set A

Key formula for non-mutually exclusive events:

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

The subtraction of P(A ∩ B) corrects for double-counting outcomes that belong to both events.

When using Venn diagrams:

1. Start by filling in the intersection region (A ∩ B) if known
2. Calculate the "A only" region: n(A) - n(A ∩ B)
3. Calculate the "B only" region: n(B) - n(A ∩ B)
4. Calculate elements outside both circles: Total - [n(A only) + n(B only) + n(A ∩ B)]
5. Calculate probabilities by dividing regions by the total

Probability from data and two-way tables

CIE questions often present data in tables and require probability calculations:

	Male	Female	Total
Passed	45	62	107
Failed	13	18	31
Total	58	80	138

From this table: P(student is male) = 58/138 = 29/69, P(student passed) = 107/138, P(female who passed) = 62/138 = 31/69.

Be careful to identify the correct denominator:

• For overall probability, use the grand total
• For conditional probability (e.g., "given the student is female"), use the total for that condition as the denominator

Worked examples

Example 1: Tree diagram without replacement

Question: A box contains 7 milk chocolates and 5 dark chocolates. Sarah selects two chocolates at random without replacement. Calculate the probability that:

(a) both chocolates are milk chocolate

(b) the two chocolates are different types

Solution:

(a) Draw a tree diagram:

First selection: P(Milk) = 7/12, P(Dark) = 5/12

Second selection:

• After selecting Milk first: P(Milk) = 6/11, P(Dark) = 5/11
• After selecting Dark first: P(Milk) = 7/11, P(Dark) = 4/11

P(both Milk) = P(Milk then Milk) = 7/12 × 6/11 = 42/132 = 7/22 [2 marks]

(b) P(different types) = P(Milk then Dark) + P(Dark then Milk)

= (7/12 × 5/11) + (5/12 × 7/11)

= 35/132 + 35/132

= 70/132 = 35/66 [3 marks]

Alternative method: P(different) = 1 - P(same type)

P(same) = P(both Milk) + P(both Dark) = 7/22 + (5/12 × 4/11) = 7/22 + 20/132 = 42/132 + 20/132 = 62/132

P(different) = 1 - 62/132 = 70/132 = 35/66

Example 2: Venn diagram and set notation

Question: In a group of 50 students, 32 study French (F), 25 study Spanish (S), and 8 study neither language.

(a) Complete a Venn diagram to show this information.

(b) Find P(F ∩ S)

(c) Find P(F ∪ S)'

Solution:

(a) Students studying at least one language = 50 - 8 = 42

Using n(F ∪ S) = n(F) + n(S) - n(F ∩ S):

42 = 32 + 25 - n(F ∩ S)

n(F ∩ S) = 57 - 42 = 15

Venn diagram regions:

• F only: 32 - 15 = 17
• S only: 25 - 15 = 10
• F ∩ S: 15
• Neither: 8 [3 marks]

(b) P(F ∩ S) = 15/50 = 3/10 [1 mark]

(c) P(F ∪ S)' = P(neither language) = 8/50 = 4/25 [1 mark]

Example 3: Expected frequency

Question: The probability that a biased coin shows heads is 0.65. The coin is tossed 200 times.

(a) Calculate the expected number of heads.

(b) The coin actually shows heads 142 times. Calculate the relative frequency of heads.

Solution:

(a) Expected frequency = probability × number of trials

= 0.65 × 200 = 130 heads [2 marks]

(b) Relative frequency = 142/200 = 0.71 [1 mark]

The relative frequency (0.71) differs from the theoretical probability (0.65), which is normal in practical experiments.

Common mistakes and how to avoid them

• Mistake: Adding probabilities when multiplication is required. Students calculate P(A and B) as P(A) + P(B) instead of P(A) × P(B) for independent events. Correction: Use the "and" rule (multiply) for events occurring together; use the "or" rule (add) only for mutually exclusive events.

• Mistake: Not adjusting probabilities in "without replacement" scenarios. After drawing one item, students use the original probability instead of updating both the favourable outcomes and total. Correction: Reduce both numerator and denominator appropriately after each selection. If 3 red balls from 10 total are drawn, the next probability uses 2 red from 9 total, not 2 from 10.

• Mistake: Confusing P(A ∪ B) and P(A ∩ B). Students use the wrong symbol or calculate the wrong probability. Correction: Remember ∪ means "union" (or), ∩ means "intersection" (and). P(A ∪ B) is larger than either individual probability; P(A ∩ B) is smaller.

• Mistake: Forgetting to subtract P(A ∩ B) in the addition rule for non-mutually exclusive events. Students write P(A ∪ B) = P(A) + P(B) even when events can occur together. Correction: Use P(A ∪ B) = P(A) + P(B) - P(A ∩ B) unless explicitly told events are mutually exclusive.

• Mistake: Using incorrect denominators for conditional probability. When asked "probability given that...", students use the total sample space instead of the restricted space. Correction: For P(A|B), the denominator is the total for condition B, not the overall total.

• Mistake: Not simplifying fractions or giving probability greater than 1. Final answers should always be simplified and checked against 0 ≤ P ≤ 1. Correction: Cancel common factors in probability fractions. If calculation gives probability > 1, recheck your working.

Exam technique for Probability

• Show clear working for multi-step calculations. CIE mark schemes award method marks even if the final answer is incorrect. Write out tree diagrams fully with probabilities on branches, show multiplication along paths, and indicate which paths you're adding. Questions worth 3+ marks typically award 1-2 method marks.

• Recognize command words. "Calculate" or "Find" requires a numerical answer with working. "Write down" suggests the answer is immediate from given information. "Show that" requires you to demonstrate the given result through clear steps. "Hence" means use your previous answer. "Complete the diagram" requires filling in specific values or probabilities, usually for 1 mark per entry.

• Check whether simplification is required. Questions asking for probability as a fraction expect simplified answers unless stated otherwise ("give your answer as a fraction in its simplest form"). Decimal answers should typically be given to 3 significant figures unless specified. Percentage answers should be stated with the % symbol.

• Use complementary probability strategically. For "at least one" questions, calculating P(none) and subtracting from 1 is usually faster than finding multiple paths on a tree diagram. Questions phrased as "at least one", "one or more", or "not all" signal this approach.

Quick revision summary

Probability measures likelihood from 0 to 1 using favourable outcomes divided by total outcomes. Multiply probabilities for independent events occurring together ("and"); add for mutually exclusive events ("or"). Tree diagrams organize sequential events—multiply along branches, add across paths. Without replacement changes subsequent probabilities. Venn diagrams use P(A ∪ B) = P(A) + P(B) - P(A ∩ B). Expected frequency equals probability times trials. Always check 0 ≤ P ≤ 1 and simplify fractions. Complementary probability P(A') = 1 - P(A) simplifies "at least" questions.