CAMBRIDGE O LEVEL · 2025–2027

Probability of combined events.

Organise combined outcomes with sample spaces, Venn diagrams and probability trees.

Syllabus 8.3Visual notesWorked examples

BY THE END

Use union and intersection · multiply along trees and add routes · distinguish replacement and independence

List outcomes in a sample space

Two random actions create pairs of results. Listing those pairs in an organised grid prevents missing possibilities and shows how different routes can satisfy the same combined event.

For two fair dice, the first die chooses a row and the second chooses a column. The 36 cells are equally likely even though the possible totals are not. Decide whether an event includes a pair before counting it, and keep ordered pairs distinct when the experiment distinguishes them.

For combined events, organise every possible outcome before counting. A table is useful for two dice, two spinners, or a coin and a die.

TWO FAIR DICE · SELECT AN EVENT

Rows: first die. Columns: second die. Each cell is an ordered pair.
↓ / →	1	2	3	4	5	6
1	(1, 1)	(1, 2)	(1, 3)	(1, 4)	(1, 5)	(1, 6)
2	(2, 1)	(2, 2)	(2, 3)	(2, 4)	(2, 5)	(2, 6)
3	(3, 1)	(3, 2)	(3, 3)	(3, 4)	(3, 5)	(3, 6)
4	(4, 1)	(4, 2)	(4, 3)	(4, 4)	(4, 5)	(4, 6)
5	(5, 1)	(5, 2)	(5, 3)	(5, 4)	(5, 5)	(5, 6)
6	(6, 1)	(6, 2)	(6, 3)	(6, 4)	(6, 5)	(6, 6)

P(sum is 7) = 636 = 16

All 36 ordered pairs are equally likely. (1, 6) and (6, 1) are different outcomes.

WORKED EXAMPLE

Two fair dice are rolled. Find P(total of 7).

The six favourable pairs are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1).

P(total of 7) = 636 = 16

Pause and explainWhy are (2, 5) and (5, 2) separate outcomes for two labelled dice?

They assign different values to the first and second dice, even though both totals are 7.

Read ‘and’ and inclusive ‘or’ in a Venn diagram

Some events overlap: a student can study both Art and Biology. A Venn diagram separates the overlap from the regions belonging to only one event, making “and” and “or” precise.

Intersection means both conditions hold. Union means at least one holds, including both. When adding event probabilities, the overlap is counted in each event total, so it must be subtracted once to obtain the union.

A ∩ B means both A and B. A ∪ B means A or B or both.

VENN DIAGRAM · 40 EQUALLY LIKELY PEOPLE

A only 10, both 8, B only 12, neither 10. Selected event A ∪ B: in A or B or both.

P(A ∪ B) = 3040 = 34

Shaded regions mean in A or B or both. The total includes the 10 people outside both circles.

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

WORKED EXAMPLE

Use the diagram to calculate the probability of A or B.

P(A ∪ B) = 1840 + 2040 − 840 = 3040 = 34

The overlap is included in both P(A) and P(B), so subtract it once.

Pause and explainWhy subtract the overlap once when adding P(A) and P(B)?

It was included twice in the two totals, but must appear only once in the probability of their union.

Separate mutually exclusive and independent

“Cannot happen together” and “does not affect the other chance” describe different relationships. The first is mutual exclusivity; the second is independence.

Mutually exclusive events have no overlap, so their probabilities add directly for an “or” event. Independent events may occur together, and their original probabilities multiply for an “and” event. Check the relationship before selecting a rule; the words are not interchangeable.

Mutually exclusive

Events cannot happen together: P(A ∩ B) = 0. On one die roll, ‘1’ and ‘6’ are mutually exclusive.

P(A ∪ B) = P(A) + P(B)

Independent

Knowing one event happened does not change the probability of the other. Separate fair coin tosses are independent.

P(A ∩ B) = P(A) × P(B)

Pause and explainAre heads and tails on one coin toss independent events?

No. They are mutually exclusive: observing heads rules out tails on that same toss.

Multiply along a branch; add alternative routes

A probability tree records an experiment one stage at a time. Each branch starts from the situation created by the earlier result, so its probability must describe that particular situation.

Without replacement, one object has gone and the remaining composition may have changed. With replacement, the original composition is restored. Multiply along one complete route because every step must occur; add separate successful routes when either route meets the event.

At every split, branch probabilities sum to 1. Write probabilities beside branches and outcomes at their ends. Multiply along a complete route to find its probability. Add probabilities of distinct routes that satisfy the question.

TREE DIAGRAM · BAG OF 3 RED AND 2 BLUE COUNTERS

Draw two counters at random. Keep the first counter out before the second draw. R = red; B = blue. Scroll the diagram sideways on a narrow screen.

Two-draw tree without replacement. First probabilities 3 fifths red, 2 fifths blue. After red: 2 over 4 red, 2 over 4 blue. After blue: 3 over 4 red, 1 over 4 blue.

P(RR) = 35 × 24 = 310
P(different colours) = 35 × 24 + 25 × 34 = 35

P(RR) = 0.30; P(different colours) = 0.60. The second probabilities depend on the first colour. There are only 4 counters left.

WORKED EXAMPLE

From 3 red and 2 blue counters, draw two without replacement. Find P(at least one blue).

Use the complement: the only way to have no blue is RR.

P(at least one blue) = 1 − P(RR)
= 1 − (35 × 24) = 710

Alternatively add RB, BR and BB. ‘Exactly one blue’ would include only RB and BR.

Pause and explainWhy is the second denominator smaller when drawing without replacement?

One object has been removed, so fewer objects remain available for the second selection.

Check that every outcome is accounted for

The final routes of a complete tree list mutually exclusive ways the experiment can finish. Exactly one final route occurs, so the route probabilities must sum to 1.

Use this as a check after multiplying branch probabilities. Also check that probabilities leaving every individual split sum to 1. These local and overall totals can reveal a wrong replacement denominator or a missing outcome before you answer the requested event.

WORKED EXAMPLE

Check the four paths for the tree without replacement.

P(RR) = 620, P(RB) = 620
P(BR) = 620, P(BB) = 220
Total = 2020 = 1

With replacement the four numerators are 9, 6, 6 and 4 over 25, again summing to 1.

Pause and explainIf complete final-route probabilities sum to 0.9, what should you suspect?

An omitted outcome or an incorrect branch calculation. The total must be 1 for all possible final routes.

Your turn. Choose a diagram and show your routes.

Choose a representation suited to the experiment: a grid for paired equally likely outcomes, a Venn diagram for overlapping events, or a tree for stages.

Translate “at least”, “exactly” and “not” carefully. Mark all successful cells, regions or routes first, then calculate and check the final probability’s range.

Two fair dice are rolled. Find P(a double).

Reveal worked answer

The six pairs (1, 1) to (6, 6) give P = 636 = 16.

Two fair dice are rolled. Find P(at least one 6).

Reveal worked answer

There are 6 pairs in the row for 6 and 6 in the column for 6, but (6, 6) is counted twice. P = (6 + 6 − 1)36 = 1136.

P(A) = 0.45, P(B) = 0.5 and P(A ∩ B) = 0.2. Find P(A ∪ B).

Reveal worked answer

0.45 + 0.5 − 0.2 = 0.75.

A bag has 3 red and 2 blue counters. Two are chosen without replacement. Find P(both blue).

Reveal worked answer

25 × 14 = 220 = 110.

For the same bag, replace the first counter before choosing the second. Find P(one counter of each colour).

Reveal worked answer

P(RB) + P(BR) = 35 × 25 + 25 × 35 = 1225.

A fair coin is tossed and a fair die rolled independently. Find P(heads and an even number).

Reveal worked answer

12 × 36 = 14. There are 3 favourable outcomes out of 12 equally likely coin–die pairs.