Tree diagrams without replacement
Update remaining counts after the first selection and include every valid order.
Extended only.
Before you begin
- Use replacement trees.
- Subtract selected items from counts.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
After one draw without replacement from five counters, how many remain?
What you will learn
- Change second-stage probabilities according to the first draw.
- Calculate combined events without replacement.
The first choice changes the second collection
With three red and two blue counters, a first red leaves two red and two blue among four. The second-stage probabilities on that branch are each 1/2. A first blue instead leaves three red and one blue, giving second-stage values 3/4 and 1/4.
Keep the denominator four on both second-stage branches, but adjust the relevant numerator separately. P(two red) = 3/5 × 2/4 = 3/10. Using 3/5 again incorrectly assumes replacement and overestimates this event.
Different branches still give a complete experiment
Exactly one red has paths RB and BR. Their products are (3/5)(2/4) = 3/10 and (2/5)(3/4) = 3/10, so the total is 3/5. If colour counts differ, the path probabilities may still be equal for one-of-each, but establish them from the branch values rather than guessing.
Use the complement for at least one red: P(BB) = (2/5)(1/4) = 1/10, so the answer is 9/10. Every node's outgoing probabilities add to one. A correct tree records the changing condition at each stage, not just a final fraction.
Pause and explain
Three red and two blue; two draws without replacement. Find P(two red).
Worked example
Three red and two blue counters are drawn twice without replacement. Find P(exactly one red).
Show the worked solution
- RB has probability (3/5)(2/4) = 3/10.
- BR has probability (2/5)(3/4) = 3/10.
- Add the two orders to obtain 3/5.
Answer 3/5.
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.
From this bag without replacement, find P(two blue).
Give me a hint
After a blue there is one blue among four.
Compare my reasoning
- The first blue probability is 2/5.
- The next blue probability is 1/4.
- The product is 1/10.
1/10.
Look for these in your work
- Updated the blue count.
- Updated the remaining total.
There are four red and three blue counters. Draw two without replacement. Find P(exactly one blue).
Give me a hint
Include RB and BR with the appropriate second counts.
Compare my reasoning
- RB probability is (4/7)(3/6) = 2/7.
- BR probability is (3/7)(4/6) = 2/7.
- The sum is 4/7.
4/7.
Look for these in your work
- Used both orders.
- Changed denominators to six.
Three red and two blue are drawn twice without replacement. Find P(at least one red) using a complement.
Give me a hint
The only complement is BB.
Compare my reasoning
- P(BB) = (2/5)(1/4) = 1/10.
- At least one red includes RR, RB and BR.
- Subtract the complement to obtain 9/10.
9/10.
Look for these in your work
- Identified all included outcomes.
- Used a changing-probability complement.
Common mistakes
- Reducing the denominator but not the selected colour count.
- Treating the draws as independent.
- Missing an alternative order.
What can you explain now?
Are consecutive colour draws without replacement independent?
Compare with the explanation
Generally no.
The first colour changes the composition and therefore the second colour probabilities.
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, compare replacement and non-replacement trees using the same starting bag.
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.
Original StudyForA teaching content · AI-assisted checks · Human teacher review pending.
Number contains 19 sequenced lessons; Algebra and graphs 21; Coordinate geometry 10; Geometry 15; Mensuration 13; Trigonometry 13; Transformations and vectors 11; Probability 7; Statistics 13. All 72 syllabus section entries now link to teaching and staged practice. Seven mixed checkpoints sample skills and suggest review lessons. Nine earlier overviews remain available. Full cumulative assessment and human teacher review remain pending. Written lesson practice, charts and constructions are self-checked, not automatically graded. Checkpoints do not save an exam grade or certify mastery.
Check the official syllabus 2025–2027 Open related practice and resources