Combine events and calculate probability
Use event notation, complements, addition and multiplication rules, conditional probability and tree diagrams.
Section Paper 2 · Written · calculator required
Exam scope and assessment notes
This lesson maps the Combine events and calculate probability section of Cambridge IGCSE Statistics 0479 (2027 and 2028–2030 syllabus versions). This is original StudyForA teaching and self-check practice, not Cambridge material, an official mark scheme, a timed mock or an official grade. Charts and written working are not automatically assessed; human subject-teacher review remains pending. Use the syllabus version for your exam year. Cambridge offers 0479 in the June series only and does not make it available in every administrative zone. Check the current Cambridge 0479 page and confirm your centre's entry eligibility with your exams officer.
Before you begin
- Use the earlier 0479 ideas where needed; write units, working and a contextual interpretation for each result.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A fair six-sided die is rolled. What is the probability of a 2?
What you will learn
- Use probability notation for events and their complements
- Calculate simple probabilities with and without replacement
- Apply the complement and addition rules
- Distinguish mutually exclusive from independent events
- Calculate simple conditional probabilities and use tree diagrams
Name events before calculating
P(A) is the probability of A; P(A′) is the probability that A does not happen. P(A ∩ B) means A and B; P(A ∪ B) means A or B or both. Probabilities lie between 0 and 1.
Use the correct relationship
P(A′) = 1 − P(A). For two events, P(A ∪ B) = P(A) + P(B) − P(A ∩ B); if they are mutually exclusive, the intersection is zero. Independent events do not change each other's probability, so P(A ∩ B) = P(A)P(B). Do not confuse independence with mutual exclusion.
Condition on what is already known
Conditional probability asks for a probability within a restricted group or after an event is known. Calculate it from the relevant counts or remaining outcomes; the syllabus does not require the formula P(A|B)=P(A∩B)/P(B). Tree diagrams show sequential choices: multiply along a path and add probabilities for alternative paths that satisfy the event.
Replacement changes the next probability
With replacement, the original composition is restored before the next draw, so probabilities can remain the same. Without replacement, counts and probabilities change after each draw. Label each branch with the probability at that stage and check each branch pair sums to 1.
Pause and explain
Two events cannot happen together. Are they independent?
Worked example
A bag has 3 red and 2 blue counters. Two are selected without replacement. Find the probability both are red.
Show the worked solution
- First red: 3/5.
- After a red counter is removed, 2 red remain among 4 counters: 2/4.
- Multiply along the path: 3/5 × 2/4 = 3/10.
Answer The probability is 3/10. Without replacement, the second probability changes after the first red counter is removed.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A student chooses one card at random from cards numbered 1 to 10. Find the probability that the number is divisible by 2 or by 5.
Give me a hint
Count the union once; number 10 is in both groups.
Compare my reasoning
- Divisible by 2: {2,4,6,8,10}, five outcomes.
- Divisible by 5: {5,10}, two outcomes; overlap is {10}.
- Union count = 5 + 2 − 1 = 6, so probability = 6/10 = 3/5.
The probability is 3/5. Outcome 10 is counted once in the union.
Look for these in your work
- Both events identified
- Overlap removed once
- Probability simplified
A box contains 4 green and 6 yellow counters. Find the probability of selecting two green counters with replacement, then without replacement.
Give me a hint
Write the first probability and update the second branch only without replacement.
Compare my reasoning
- With replacement: (4/10)(4/10) = 4/25.
- Without replacement: (4/10)(3/9) = 2/15.
- The outcomes differ because the first counter is returned in only one experiment.
With replacement, 4/25; without replacement, 2/15.
Look for these in your work
- Replacement case correct
- No-replacement count updated
- Difference explained
A weather forecast says the probability of rain is 0.3. A fair spinner has a 1/4 chance of blue and is independent of the weather. Find the probability of rain and blue, and explain why independence matters.
Give me a hint
For independent events, multiply the two probabilities.
Compare my reasoning
- Convert or keep consistent forms: 0.3 × 1/4.
- 0.3 × 0.25 = 0.075.
- Independence permits the product; if the events were dependent, the second probability could change.
The probability is 0.075. The multiplication rule applies because rain and the spinner result are stated to be independent.
Look for these in your work
- Probability product correct
- Decimal correct
- Independence condition stated
Common mistakes
- Adding probabilities for independent events when finding ‘and’
- Multiplying probabilities for mutually exclusive alternatives
- Leaving replacement status out of a repeated-draw calculation
- Treating ‘or’ as always mutually exclusive
What can you explain now?
If P(A)=0.35, what is P(A′)?
Compare with the explanation
0.65
The event and its complement have total probability 1.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
How does ‘mutually exclusive’ differ from ‘independent’?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.