Probability

Summary: Covers basic probability, sample space diagrams, tree diagrams (with and without replacement), conditional probability, Venn diagrams for probability, expected frequency, and relative frequency. Probability appears in both Core and Extended IGCSE Mathematics 0580 papers. Tags: igcse mathematics probability Created: 2026-07-16 Last Updated: 2026-07-16

1. Basic Probability

Probability of an event: P(event) = number of favourable outcomes / total number of possible outcomes

The probability scale: Probabilities range from 0 (impossible) to 1 (certain).

• P(certain event) = 1
• P(impossible event) = 0
• The sum of probabilities of all mutually exclusive possible outcomes = 1

Complementary events: P(not A) = 1 − P(A)

Example: A fair six-sided die is rolled. Find: (a) P(rolling a 4) = 1/6 (b) P(rolling an even number) = 3/6 = 1/2 (c) P(not rolling a 6) = 1 − 1/6 = 5/6

Mutually exclusive events: Two events cannot happen at the same time.

• Addition rule: P(A or B) = P(A) + P(B)

Not mutually exclusive: P(A or B) = P(A) + P(B) − P(A and B)

Example: A bag has 3 red, 4 blue, 5 green balls. One is drawn. Find P(red or blue). P(red or blue) = 3/12 + 4/12 = 7/12.

2. Independent Events

Two events are independent if the outcome of one does not affect the outcome of the other.

Multiplication rule: P(A and B) = P(A) × P(B)

Example: The probability that Alice passes Maths is 0.7 and English is 0.8. Assuming independence, find P(she passes both). P(both) = 0.7 × 0.8 = 0.56.

3. Sample Space Diagrams

A sample space is the set of all possible outcomes. Represent it systematically using lists or two-way tables.

Example: Two fair coins are flipped. List the sample space and find P(at least one head). Sample space = {HH, HT, TH, TT} (4 equally likely outcomes). P(at least one head) = 3/4 = 0.75.

Two-way table: Rolling two dice and adding the scores produces a 6×6 table (36 equally likely outcomes). P(total = 7) = 6/36 = 1/6 (the pairs {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)} all sum to 7).

4. Tree Diagrams

Tree diagrams show the outcomes of successive events, with probabilities on each branch.

Key rules:

• Probabilities on branches from the same point sum to 1.
• Multiply along branches for combined outcomes (AND).
• Add probabilities of relevant final outcomes (OR).
• The sum of all final outcome probabilities = 1.

Without replacement: Probabilities change after each event because the total number of items decreases. The second set of branches shows conditional probabilities.

With replacement: Probabilities stay the same because items are returned.

Example (without replacement): A bag has 4 red and 6 blue counters. Two are drawn without replacement. Find P(both red). First draw: P(red) = 4/10 = 2/5. Second draw (given first was red): P(red) = 3/9 = 1/3. P(both red) = 2/5 × 1/3 = 2/15.

Example (with replacement): Same bag, but counters are replaced. Find P(both red). First: P(red) = 4/10 = 2/5. Second: P(red) = 4/10 = 2/5. P(both red) = 2/5 × 2/5 = 4/25.

Example (exactly one): Box A has 2 red and 3 green sweets. Two are picked without replacement. Find P(exactly one is red). P(R then G) = 2/5 × 3/4 = 6/20. P(G then R) = 3/5 × 2/4 = 6/20. P(exactly one red) = 6/20 + 6/20 = 12/20 = 3/5.

5. Conditional Probability (Extended)

Conditional probability P(A|B) is the probability of A given that B has occurred: P(A|B) = P(A and B) / P(B)

Example: In a class, 12 study French (F), 8 study German (G), 4 study both. A student chosen at random studies French. Find P(this student also studies German). P(G|F) = P(F and G) / P(F) = (4/total) / (12/total) = 4/12 = 1/3.

Tree diagrams and conditional probability: The second set of branches in a tree diagram represents conditional probabilities given the outcome of the first event. This is why “without replacement” problems involve conditional probabilities on the second tier.

6. Venn Diagrams for Probability

The rectangle represents the sample space (total probability = 1). The probability of an event is proportional to the area of its region.

Example: P(A) = 0.4, P(B) = 0.5, P(A ∩ B) = 0.2. Find: (a) P(A ∪ B) = 0.4 + 0.5 − 0.2 = 0.7 (b) P(A only) = 0.4 − 0.2 = 0.2 (c) P(neither) = 1 − 0.7 = 0.3

7. Expected Frequency

Expected frequency = probability × number of trials

Example: The probability a biased coin lands on heads is 0.35. The coin is flipped 200 times. Expected heads = 0.35 × 200 = 70.

8. Relative Frequency

Relative frequency = number of times event occurs / total number of trials

As the number of trials increases, relative frequency tends towards the theoretical probability (the Law of Large Numbers). Use relative frequency to estimate probability when the theoretical probability is unknown.

Example: A spinner is spun 150 times and lands on blue 42 times. Estimate P(blue). Estimated P(blue) = 42/150 = 0.28 (or 14/75).

Example: A biased coin was tossed 80 times and came up heads 28 times. Estimate how many heads you would expect in 500 tosses. Estimated P(heads) = 28/80 = 0.35. Expected in 500 tosses = 0.35 × 500 = 175 heads.

Sources

• BBC Bitesize GCSE Mathematics — Probability and Tree Diagrams, BBC (free educational resource)
• OpenStax Math — Probability and Statistics, Rice University (free, CC BY 4.0)
• Cambridge IGCSE Mathematics 0580 — Probability, Cambridge Assessment International Education
• CK-12 Mathematics — Probability, CK-12 Foundation (free, CC BY-NC 3.0)

Related Notes

• Number — Fractions, decimals, and percentages used in probability
• Statistics — Relative frequency links probability to statistics
• Algebra and Graphs — Tree diagrams as graph-like structures
• IGCSE-Maths-Index — Full IGCSE Mathematics index

Common Misconceptions