Combined Events and Conditional ProbabilityEdexcel GCSE Maths: Revision notes

Section 1

What are mutually exclusive and independent events?

Mutually exclusive events cannot occur at the same time. If one happens, the other cannot. For example, rolling a dice and getting a 3 or a 5 are mutually exclusive.

Independent events occur when the probability of one event is not affected by whether the other occurs. Rolling two dice gives independent events because the first roll doesn't change the probability of the second.

Property	Mutually Exclusive	Independent
Can both occur simultaneously?	No	Yes
Does one affect the other's probability?	Yes (if one happens, other cannot)	No
Addition rule	P(A or B) = P(A) + P(B)	P(A or B) = P(A) + P(B) − P(A and B)
Multiplication rule	P(A and B) = 0	P(A and B) = P(A) × P(B)

For mutually exclusive events, if they also cover all possibilities, their probabilities sum to 1.

Key termsmutually exclusive eventsindependent events

Common mistake

Students often confuse independent with mutually exclusive. Remember: independent events CAN happen together (like rolling two 6s), while mutually exclusive events CANNOT (like rolling a 3 AND a 5 on one dice).

Think of it like this

Independent events are like two separate coin flips – one person's result doesn't affect another's. Mutually exclusive events are like voting for one candidate – choosing one option removes the others.

Section 2

How do you use the addition rule for combined events?

The addition rule finds the probability that event A or event B (or both) occurs:

P(A or B) = P(A) + P(B) − P(A and B)

The term P(A and B) is subtracted because it's counted twice when you add P(A) and P(B). This overlap is called the intersection.

For mutually exclusive events where P(A and B) = 0:

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

Example: Finding probability with overlap

P(drawing a red card) = 26/52
P(drawing a King) = 4/52
P(drawing a red King) = 2/52
P(red card OR King) = 26/52 + 4/52 − 2/52 = 28/52 = 7/13

Without subtracting the overlap, you'd count the two red Kings twice.

Key termsaddition ruleintersectionunion

Exam tip

Always check if the events are mutually exclusive first. If they are, you don't need the subtraction term. Examiners look for correct use of this simplification.

Example

A spinner has outcomes: 1, 2, 3, 4, 5. Event A = {1, 2, 3}, Event B = {3, 4, 5}. P(A) = 3/5, P(B) = 2/5, P(A and B) = 1/5. Therefore P(A or B) = 3/5 + 2/5 − 1/5 = 4/5.

Section 3

How do you apply conditional probability?

Conditional probability is the probability of an event occurring given that another event has already happened. Written as P(B|A), meaning "the probability of B given that A has occurred".

Formula: P(B|A) = P(A and B) / P(A)

Rearranged: P(A and B) = P(A) × P(B|A) (the multiplication rule)

For independent events, P(B|A) = P(B) because one event doesn't affect the other.

Example: Drawing without replacement

A bag has 4 red and 3 blue balls
P(first ball is red) = 4/7
P(second is blue | first was red) = 3/6 = 1/2 (only 3 blue left, 6 balls total)
P(red then blue) = 4/7 × 1/2 = 4/14 = 2/7

Notice the denominator changes because we're sampling without replacement. This makes events dependent.

Key termsconditional probabilitymultiplication ruledependent events

Exam tip

When writing out solutions for conditional probability, always clearly state what has been given (the condition). Show the reduced sample space or explain why the probability changes.

Common mistake

Students often forget to update probabilities when sampling without replacement. After removing an item, both the number of favourable outcomes AND the total number of outcomes change.

Section 4

How do you use tree diagrams for combined events?

Tree diagrams visually show all possible outcomes of combined events and their probabilities.

For independent events:

The probability on second branches stays the same regardless of the first outcome
Example: Flipping two fair coins – each branch is 1/2

For dependent events:

The probabilities on second branches change based on the first outcome
Example: Drawing without replacement – the second probability depends on what was drawn first

Rules for tree diagrams:

At each branch point, probabilities must sum to 1
Multiply along branches to find the probability of a path
Add paths to find the probability of a combined event

Example calculation:

First draw red (4/7), then blue (3/6): probability = 4/7 × 3/6 = 12/42
First draw blue (3/7), then red (4/6): probability = 3/7 × 4/6 = 12/42
P(one red and one blue) = 12/42 + 12/42 = 24/42 = 4/7

Key termstree diagrambranchespaths

Exam tip

Show your working by writing the probability calculation on each branch. Examiners want to see whether you've correctly identified dependent or independent events before you draw the diagram.

Example

A bag has red and blue balls. Draw one, replace it, draw another. Both draws have P(red) = 2/5 and P(blue) = 3/5 on every branch (independent). Draw one, don't replace, draw another: the second branch probabilities depend on the first outcome (dependent).

Section 5

How do you represent events using Venn diagrams and set notation?

Venn diagrams show events as overlapping regions, making it easy to visualise combined and conditional probabilities.

Key notation:

A ∪ B (union): elements in A or B or both
A ∩ B (intersection): elements in both A and B
A' (complement): elements not in A
ξ (xi): the universal set (all possible outcomes)

Reading probabilities from Venn diagrams:

P(A) = sum of all regions inside A
P(A and B) = sum of regions in the overlap
P(A or B) = sum of all regions in A or B
Regions outside both circles represent P(neither A nor B)

Setting up a Venn diagram:

Identify the universal set size
Fill in the intersection first (items in both A and B)
Fill remaining parts of each circle
Fill outside region (items in neither)
Check that all regions sum to the universal set total

Example with 100 students:

30 study Maths, 25 study Physics, 10 study both
Maths only: 30 − 10 = 20
Physics only: 25 − 10 = 15
Neither: 100 − 20 − 15 − 10 = 55
P(Maths or Physics) = (20 + 10 + 15)/100 = 45/100 = 9/20

Key termsVenn diagramunionintersectioncomplementset notation

Exam tip

Always label regions in Venn diagrams clearly with their values or probabilities. Examiners mark whether you've correctly placed information in the intersection versus the non-overlapping parts.

Section 6

How do you solve probability problems using systematic listing?

Systematic listing ensures you find all possible outcomes for combined events without missing any or double-counting.

Strategies:

List outcomes in order – arrange outcomes in a logical sequence (e.g., smallest to largest)
Use tables – particularly useful for two independent events (create rows and columns)
Use tree diagrams – show all branches systematically, left to right
Count carefully – total the outcomes that satisfy your condition

Example: Two fair dice

Find P(total of 7):

Outcomes: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
That's 6 outcomes
Total possible outcomes = 36
P(total 7) = 6/36 = 1/6

Example: Three coins

Find P(at least 2 heads):

List: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
Count with at least 2 heads: HHH, HHT, HTH, THH = 4 outcomes
P(at least 2 heads) = 4/8 = 1/2

Common pitfalls:

Missing outcomes by not being systematic
Miscounting the total number of outcomes
Forgetting whether order matters (e.g., (1,2) and (2,1) are different outcomes when using two dice)

Key termssystematic listingsample spaceexhaustiveoutcome space

Exam tip

Show your systematic list in your working. Examiners want evidence you've found all outcomes methodically. A clearly organised list is worth marks even if your final answer is slightly wrong.

Common mistake

Students often treat (1,2) and (2,1) as one outcome when listing results from two dice or two events. They are separate outcomes because order matters – clearly distinguish them in your list.

Must Know

Mutually exclusive events cannot both occur; their probabilities add directly: P(A or B) = P(A) + P(B). Independent events are unaffected by each other; P(A and B) = P(A) × P(B).
The addition rule: P(A or B) = P(A) + P(B) − P(A and B). The subtraction corrects for the overlap (intersection) being counted twice.
Conditional probability: P(B|A) means "probability of B given A has occurred". Use the formula P(B|A) = P(A and B) / P(A), or rearranged: P(A and B) = P(A) × P(B|A).
Tree diagrams show dependent events by changing branch probabilities based on previous outcomes. Always multiply along a path, then add paths for combined events.
Venn diagrams with set notation (∪ for union, ∩ for intersection, ' for complement) clearly show how events overlap and make calculating combined probabilities visual and systematic.
Systematic listing of all possible outcomes (in tables, lists, or tree diagrams) ensures no outcomes are missed and prevents double-counting in probability calculations.

That's the notes covered.

Carry on to the next subtopic.