Calculate probabilities of combined events
Organise combined outcomes with sample spaces, Venn diagrams and tree diagrams, then add or multiply along the correct paths.
Use this pathway for Cambridge IGCSE International Mathematics 0607, 2028–2030. Core and Extended objectives are labelled in each lesson; Extended adds requirements beyond Core. Paper 1 and Paper 2 are non-calculator. Papers 3–6 require a graphic display calculator (GDC). These original lessons use self-checked practice; they are not a Cambridge grade, full timed paper or teacher review.
Before you begin
- Check whether your school has entered you for Core or Extended. Core and Extended share most Number ideas, but some notation and methods are Extended only.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
In a tree, how do you calculate the probability of one complete path?
What you will learn
- Represent combined events with sample spaces, two-set Venn diagrams and trees
- Calculate probabilities with and without replacement
- Distinguish mutually exclusive outcomes from independent events
Choose a representation that shows every outcome
A sample space lists all possible ordered outcomes; a tree shows stages and branch probabilities; a Venn diagram groups outcomes by set membership. In Core, combined-event questions are with replacement and Venn diagrams use at most two sets.
Extended may include without-replacement trees, where the second branch probabilities change after the first result. Put outcomes at branch ends and probabilities beside each branch.
Multiply along a route and add separate routes
For a sequence, multiply the probabilities along one complete path. For an event reached by alternative non-overlapping paths, add the path probabilities.
For independent events, P(A and B) = P(A)P(B). For mutually exclusive events, P(A or B) = P(A) + P(B). These conditions are different: independence is not the same as mutually exclusive. Extended notation may use ∩ for ‘and’ and ∪ for ‘or’.
Pause and explain
If a counter is not replaced, what happens to the second-draw probabilities?
Worked example
A fair coin is tossed twice. Find the probability of exactly one head.
Show the worked solution
- The equally likely ordered outcomes are HH, HT, TH and TT.
- Exactly one head occurs in HT and TH: 2 outcomes out of 4.
- Probability = 2/4 = 1/2.
Answer 1/2.
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 fair six-sided die is rolled twice. Find the probability of a total of 7.
Give me a hint
List ordered pairs with a total of 7 out of 36 equally likely pairs.
Compare my reasoning
- The pairs are (1,6),(2,5),(3,4),(4,3),(5,2),(6,1).
- There are 6 successful pairs among 36.
- Probability = 6/36 = 1/6.
1/6.
Look for these in your work
- Ordered outcomes counted
- All 36 outcomes used
- Fraction simplified
A bag has 3 red and 2 blue counters. Two are drawn without replacement. Find the probability of red then blue.
Give me a hint
After drawing red, update the total and blue count for the second stage.
Compare my reasoning
- P(red first) = 3/5.
- After a red is removed, 2 blue counters remain among 4, so P(blue second | red first) = 2/4.
- Multiply along the path: 3/5 × 2/4 = 3/10.
3/10.
Look for these in your work
- First branch correct
- Without-replacement denominator updated
- Path probabilities multiplied
In a class, 14 learners study art, 11 study music and 6 study both. Find the probability that a randomly chosen learner from a class of 25 studies art or music, and state why the overlap is subtracted.
Give me a hint
Use P(A ∪ B) counts: combine the groups but count learners in both only once.
Compare my reasoning
- Art or music count = 14 + 11 − 6 = 19.
- Divide by the class total: 19/25.
- The six learners in both groups were counted twice in 14 + 11, so subtract them once.
19/25; subtract the overlap to count each learner once.
Look for these in your work
- Union count calculated
- Universal total used
- Overlap reason explained
Common mistakes
- Adding along a sequence instead of multiplying
- Multiplying alternative paths instead of adding
- Keeping the original probabilities after a draw without replacement
- Calling overlapping events mutually exclusive
What can you explain now?
A fair coin is tossed twice. What is P(two heads)?
Compare with the explanation
1/4
Multiply 1/2 by 1/2 for the HH path.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
When do you multiply path probabilities, and when do you add probabilities from separate successful paths?
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.