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Cambridge IGCSE Mathematics 0580 · Probability

Probability of combined events

Two or more events together: listing outcomes in a sample space diagram, multiplying and adding on tree diagrams, and reading combined events from Venn diagrams. Extended: the notation P(A∩B) and P(A∪B), and events without replacement, where the second probabilities change.

C8.3E8.3Core and ExtendedFree lesson
What the syllabus says

C8.3 (Core)

  • Calculate the probability of combined events using, where appropriate:
  • sample space diagrams
  • Venn diagrams
  • tree diagrams.

Notes and examples

  • Combined events will only be with replacement.
  • Venn diagrams will be limited to two sets.
  • In tree diagrams, outcomes will be written at the end of the branches and probabilities by the side of the branches.

E8.3 (Extended)

  • Calculate the probability of combined events using, where appropriate:
  • sample space diagrams
  • Venn diagrams
  • tree diagrams.

Notes and examples

  • Combined events could be with or without replacement.
  • The notation P(A∩B) and P(A∪B) may be used in the context of Venn diagrams.
  • On tree diagrams outcomes will be written at the end of branches and probabilities by the side of the branches.

Quoted from the Cambridge IGCSE Mathematics 0580 syllabus for 2025 to 2027.

Combined events

A combined event involves two or more things happening: two dice rolled together, a coin thrown and a spinner spun, two counters taken from a bag one after the other, or the weather on two different days.

There are three tools for working out these probabilities, and part of the skill is choosing the right one:

  • a sample space diagram, when two experiments each have a small number of equally likely outcomes;
  • a tree diagram, when the outcomes are not equally likely or there are several stages;
  • a Venn diagram, when people or items are sorted into two overlapping groups.

On the Core papers combined events are always with replacement, or independent: what happens first does not change the probabilities for what happens next. Extended questions may also be without replacement.

Sample space diagrams

A sample space diagram lists every possible outcome of two experiments in a grid: the outcomes of one experiment along the top, the other down the side, and each cell shows one combination.

If each experiment is fair, every cell is equally likely. So

probability=number of cells that fit the eventtotal number of cells

Two fair dice give 6×6=36 cells. A fair coin and a fair dice give 2×6=12. The cells can show a total, a difference, a product or just the pair of results, such as (H, 4).

Why not just list the totals? With two dice the totals 2 to 12 are not equally likely. A total of 2 happens only one way, (1, 1), but a total of 7 happens six ways. Counting cells takes this into account; counting totals does not. Note too that (2, 5) and (5, 2) are different cells: the first dice shows 2 in one and 5 in the other.

Worked example 1

Two fair six-sided dice are rolled and the scores are added. Use the sample space diagram to find the probability that the total is (a) 7, (b) more than 9, (c) an even number.

Totals when two fair dice are rolled
+123456
1234567
2345678
3456789
45678910
567891011
6789101112
  1. There are 36 equally likely cells.

  2. (a) A total of 7 appears in 6 cells: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1). Probability 636=16.

  3. (b) More than 9 means 10, 11 or 12: 3+2+1=6 cells. Probability 636=16.

  4. (c) Count the even totals row by row: there are 3 in every row, so 3×6=18 cells. Probability 1836=12.

Answer: (a) 16 (b) 16 (c) 12

Check yourself

A fair coin is thrown and a fair four-sided spinner, numbered 1 to 4, is spun. Find the probability of getting a head and an even number.

Answer it in the portal, where your answer is checked as you type →

Independent events: multiply for "and"

Two events are independent when one happening does not change the probability of the other. Two dice, a coin and a spinner, and a counter that is put back before the next is taken are all independent.

For independent events,

probability of A and B=probability of A×probability of B

Why multiply? Think of a large number of trials. A six comes up on the first dice in 16 of them. Of those trials, the second dice shows a six in 16. So both are sixes in 16 of 16 of the trials: 16×16=136, exactly what the sample space shows (one cell of 36).

For outcomes that cannot happen together, such as "red then blue" and "blue then red", the probability of one or the other is found by adding. So: multiply along a path for "and"; add different paths for "or".

Tree diagrams

A tree diagram shows each stage as a set of branches.

  • The outcomes are written at the end of each branch.
  • The probabilities are written beside each branch.
  • The branches leaving any one point cover every possibility, so their probabilities add up to 1.

To use it:

  1. Find each path that gives the event you want.
  2. Multiply the probabilities along each path.
  3. Add the results for the different paths.

When the first item is replaced, the second set of branches is the same as the first. A useful check: the probabilities at the ends of all the paths add up to 1.

Worked example 2

A bag contains 2 red counters and 3 blue counters. A counter is taken at random, its colour is noted and it is put back. A second counter is then taken. Find the probability that (a) both counters are blue, (b) the two counters are the same colour.

2/5Red2/5Red3/5Blue3/5Blue2/5Red3/5BlueFirst counterSecond counter
  1. Draw the tree. Each counter is replaced, so both stages have red 25 and blue 35.

  2. (a) Path blue then blue: 35×35=925.

  3. (b) Same colour means red then red, or blue then blue.

  4. Red, red: 25×25=425.

  5. Add the two paths: 425+925=1325.

Answer: (a) 925 (b) 1325

"At least one" and the shortcut through 1

"At least one" means one or more. With two or three stages there are many paths that have at least one success, but only one path has none. It is quicker and safer to use

probability of at least one=1−probability of none

A very common wrong method is to add the single probabilities. For two days each with probability 0.3 of rain, 0.3+0.3=0.6 is wrong, because it counts the case of rain on both days twice. With four such days, adding would give 1.2, which is impossible.

Worked example 3

The probability that it rains on any one day is 0.3, independently of other days. Find the probability that it rains on at least one of the next three days.

  1. The probability that it does not rain on a day is 1−0.3=0.7.

  2. No rain on all three days: 0.7×0.7×0.7=0.343.

  3. At least one rainy day: 1−0.343=0.657.

Answer: 0.657

Check yourself

The probability that Jo wins a game is 0.4. She plays two games, and the results are independent. Find the probability that Jo wins exactly one of the two games.

Answer it in the portal, where your answer is checked as you type →

Venn diagrams for combined events

A Venn diagram with two sets shows two events at once. Each region can hold a number of people or a probability.

  • "A and B" is the overlap.
  • "A or B" means in A, in B, or in both: everything inside either circle.
  • "A but not B" is the part of A outside the overlap.
  • "Neither" is the region outside both circles.

All four regions together make the whole: they add to the total number, or to 1 for probabilities. That is how a missing value is found. On the Core papers Venn diagrams have two sets.

Worked example 4

The Venn diagram shows the probabilities that a student chosen at random from a college studies art (A) and music (M). (a) Find x. (b) Find the probability that the student studies art or music or both.

ℰAM0.3x0.250.2
  1. (a) The four regions add to 1: 0.3+x+0.25+0.2=1, so x=1−0.75=0.25.

  2. (b) Everything inside the circles: 0.3+0.25+0.25=0.8.

  3. Or: 1−0.2=0.8, since only the outside region is excluded.

Answer: (a) 0.25 (b) 0.8

Venn diagram notation Extended

On the Extended papers, events on a Venn diagram may be written with set notation:

  • P(A∩B) is the probability of A and B: the overlap;
  • P(A∪B) is the probability of A or B (or both): everything inside the two circles;
  • P(A∩B′) is the probability of A but not B;
  • P((A∪B)′) is the probability of neither.

When you are given P(A), P(B) and P(A∩B), put the overlap in first, then subtract it to fill the "only" regions. Adding P(A)+P(B) counts the overlap twice, so

P(A∪B)=P(A)+P(B)−P(A∩B)

Worked example 5 Extended

P(A)=0.5, P(B)=0.4 and P(A∩B)=0.15. Find (a) P(A∪B), (b) P(A′∩B), (c) P((A∪B)′).

  1. Fill the Venn diagram: overlap 0.15, A only 0.5−0.15=0.35, B only 0.4−0.15=0.25.

  2. (a) P(A∪B)=0.35+0.15+0.25=0.75. Check: 0.5+0.4−0.15=0.75.

  3. (b) A′∩B is in B but not in A: 0.25.

  4. (c) Neither: 1−0.75=0.25.

  5. ℰAB0.350.150.250.25

Answer: (a) 0.75 (b) 0.25 (c) 0.25

Without replacement Extended

If the first item is not put back, the second stage is different: there is one item fewer altogether, and one fewer of the kind already taken.

Take a bag of 4 red and 6 blue counters. After a red counter is taken, 9 counters are left and only 3 are red, so the probability that the second is red is 39. After a blue counter is taken, 9 are left and 4 are still red, so it is 49.

The method on the tree is unchanged: multiply along the paths and add the paths you want. Check each pair of second branches adds to 1, such as 39+69=1.

The same applies to choosing people: picking two students from a group is always without replacement, because the same person cannot be picked twice.

More generally, Extended questions may give events that depend on each other, such as "if it rains today, the probability that it rains tomorrow is 0.6". The second set of branches then has different probabilities after each first outcome, exactly as without replacement.

Worked example 6 Extended

A bag contains 4 red counters and 6 blue counters. Two counters are taken at random without replacement. Find the probability that they are different colours.

4/10Red3/9Red6/9Blue6/10Blue4/9Red5/9BlueFirst counterSecond counter
  1. First counter: red 410, blue 610.

  2. After red: 9 left, 3 red and 6 blue. After blue: 9 left, 4 red and 5 blue.

  3. Red then blue: 410×69=2490.

  4. Blue then red: 610×49=2490.

  5. Different colours: 2490+2490=4890=815.

Answer: 815

Check yourself Extended

A bag contains 3 red sweets and 2 green sweets. Two sweets are taken at random without replacement. Find the probability that both sweets are green.

Answer it in the portal, where your answer is checked as you type →

Key points

  • Sample space diagram: count the equally likely cells; (2, 5) and (5, 2) are different cells.

  • Independent events: multiply for "and". Outcomes that cannot happen together: add for "or".

  • Tree diagrams: outcomes at the ends of branches, probabilities beside them; each set of branches adds to 1.

  • Multiply along a path, then add the paths you want.

  • At least one = 1 minus the probability of none.

  • Venn diagram: "and" is the overlap; "or" is everything in either circle.

  • Extended: P(A∪B)=P(A)+P(B)−P(A∩B).

  • Extended: without replacement, the second stage has one fewer item in total and one fewer of the kind already taken.

Where marks are lost

  • Adding probabilities along a path instead of multiplying.

  • Finding only one path, such as red then blue, when "one of each" also includes blue then red.

  • Adding the single probabilities for "at least one". Use 1 minus the probability of none.

  • Treating the possible totals of two dice as equally likely.

  • Forgetting that the branches from each point must add to 1 when completing a tree.

  • Extended: keeping the same fractions on the second branches when the item is not replaced.

  • Extended: reducing the bottom of the fraction but not the top (or the other way round) without replacement.

  • Extended: adding P(A)+P(B) for P(A∪B) and counting the overlap twice.

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An original lesson written to the Cambridge IGCSE Mathematics 0580 syllabus (2025 to 2027). The worked examples and questions are our own, not Cambridge questions. Back to every IGCSE Mathematics 0580 lesson.