IGCSE Probability Tree Diagrams
Learn probability trees for IGCSE with worked examples of draws with and without replacement.
By Math School Editorial Team
Why probability tree diagrams matter
A tree diagram organises every possible outcome when an event happens more than once. It is especially useful when a question describes two draws from a bag or two consecutive choices.
The first question to ask is whether the first event changes the probabilities of the second. Cambridge IGCSE Mathematics (0580), 2025–2027, includes tree diagrams for combined events. Core combined-event questions involve replacement; Extended questions can involve draws with or without replacement. The following are original examples, not copied exam questions.
Begin with a bag of counters
Imagine a bag containing four red counters and three blue counters. There are seven counters in total. The first-draw probability of red is 4/7 and the probability of blue is 3/7. These add to one.
Example 1: With replacement
“With replacement” means the first counter is returned before the second draw. The bag still has four red and three blue counters, so the second-draw probabilities remain 4/7 and 3/7.
To find the probability of two red counters, multiply along the red–red path:
P(red then red) = 4/7 × 4/7 = 16/49.
For exactly one red counter, two orders work:
P(red then blue) = 4/7 × 3/7 = 12/49.
P(blue then red) = 3/7 × 4/7 = 12/49.
Add them: 12/49 + 12/49 = 24/49.
The key rule is to multiply along a path and add the separate paths that satisfy the question.
Example 2: Without replacement
“Without replacement” means the first counter is kept out. Only six counters remain.
If the first counter was red, the bag now contains three red and three blue counters. The second-draw probabilities are 3/6 and 3/6.
If the first counter was blue, four red and two blue counters remain. The second-draw probabilities are 4/6 and 2/6.
For two red counters:
P(red then red) = 4/7 × 3/6 = 12/42 = 2/7.
For exactly one red counter, add the two qualifying paths:
P(red then blue) = 4/7 × 3/6 = 12/42.
P(blue then red) = 3/7 × 4/6 = 12/42.
Total = 24/42 = 4/7.
Students sometimes reduce the denominator from seven to six but forget to adjust the colour counts. The second probability must describe the counters that actually remain.
The shortcut for “at least one”
“At least one blue” includes one blue or two blues. Its opposite is no blue, meaning two reds.
Without replacement:
P(at least one blue) = 1 − P(two reds) = 1 − 2/7 = 5/7.
This is the complement method. It is often faster when the desired event involves several paths but its opposite involves only one.
Three habits that prevent mistakes
Check each split. Probabilities leaving the same point on the tree should add to one.
Read the wording carefully. Replacement keeps the composition unchanged; no replacement changes both the total and the count of one colour.
Check all final paths. In the without-replacement example, the probabilities for red–red, red–blue, blue–red and blue–blue are 12/42, 12/42, 12/42 and 6/42. Together they equal one.
Original practice questions
1. A bag contains three green and two yellow counters. One is drawn, replaced and drawn again. What is the probability that both are yellow?
2. From the same bag, two counters are drawn without replacement. What is the probability that both are yellow?
3. With no replacement, what is the probability that at least one counter is green?
Answers: (1) 2/5 × 2/5 = 4/25. (2) 2/5 × 1/4 = 1/10. (3) 1 − 1/10 = 9/10.
How parents can help
Use real coloured buttons or counters in a cup. Draw one, ask the student what remains, and have them explain the new denominator before multiplying anything. Once that is secure, practise the language “exactly one,” “at least one” and “neither.”
A diagnostic exercise can distinguish a misunderstanding of replacement from an ordinary fraction error, helping revision focus on the actual gap.
Official source: Cambridge IGCSE Mathematics 0580 syllabus 2025–2027, section 8. Verified 22 October 2026.
