Tree diagrams with replacement
Multiply along a path and add distinct final paths when trials keep the same probabilities.
Core and Extended. This is the separate 9–1 qualification 0980. Cambridge intends Core for grades 4–1 and Extended for grades 9–5; the eligible award bands are grades 5–1 for the Core papers and grades 9–3 for the Extended papers.
Before you begin
- Multiply fractions.
- Distinguish one path from an event with several paths.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
After replacement, what happens to the bag's composition?
What you will learn
- Label a two-stage probability tree.
- Find a combined event using path products and sums.
Branches show the next possible outcome
A bag contains three red and two blue counters. Draw one, replace it, then draw again. Each stage has P(red) = 3/5 and P(blue) = 2/5 because replacement restores the original collection. Write these probabilities beside the branches and the outcomes at their ends.
Multiply along a complete path: P(red then blue) = 3/5 × 2/5 = 6/25. The branches leaving any one node add to one. At the final stage, the four ordered paths RR, RB, BR and BB are distinct.
Add paths for an event containing alternatives
Exactly one red occurs on RB or BR. These paths cannot happen together in one two-draw experiment, so add their probabilities: 6/25 + 6/25 = 12/25. Two red is just RR, giving 9/25. At least one red can also be found as 1 − P(BB) = 21/25.
Multiplication is used for successive branch choices, while addition is used for disjoint complete alternatives. Core tree selections use replacement. Extended includes selection without replacement, where the second-stage branch values depend on the first outcome.
Pause and explain
Three red and two blue; two draws with replacement. Find P(exactly one red).
Worked example
A bag has three red and two blue counters. Draw twice with replacement. Find P(exactly one red).
Show the worked solution
- The paths are RB and BR.
- Each has probability (3/5)(2/5) = 6/25.
- Add the two disjoint paths to get 12/25.
Answer 12/25.
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.
Using this bag with replacement, find P(two red).
Give me a hint
There is one RR path.
Compare my reasoning
- The first red probability is 3/5.
- Replacement keeps the second red probability at 3/5.
- Multiply to obtain 9/25.
9/25.
Look for these in your work
- Kept replacement probabilities unchanged.
- Multiplied along the path.
Using the same experiment, find P(at least one red).
Give me a hint
Complement the event two blue.
Compare my reasoning
- P(BB) = (2/5)² = 4/25.
- At least one red is its complement.
- Probability is 1 − 4/25 = 21/25.
21/25.
Look for these in your work
- Selected the correct complement.
- Included both one-red and two-red cases.
Two independent events have probabilities 0.4 and 0.7. Find P(first occurs and second does not).
Give me a hint
The second non-occurrence has complementary probability.
Compare my reasoning
- P(second does not) = 0.3.
- Independence permits multiplication of 0.4 and 0.3.
- The combined probability is 0.12.
0.12.
Look for these in your work
- Used the complement.
- Stated the independence condition.
Common mistakes
- Adding probabilities along a path.
- Counting only one order for exactly one event.
- Changing branch counts despite replacement.
What can you explain now?
When do you add final tree-path probabilities?
Compare with the explanation
For disjoint complete paths satisfying the same requested event.
The paths represent alternatives that cannot occur together in one trial.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Tomorrow, draw and label all four paths for two replacement draws, then check their total probability is one.
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.