Restrict the sample space when a previous outcome is known
Read a branch probability after a known first outcome · Distinguish joint from restricted probability · Normalise the relevant path probabilities when information is given
Section E9.5 · Extended · Possibility diagrams and successive selection
Exam scope and assessment notes
Extended content; not a Core requirement. Historical study of Cambridge IGCSE Mathematics (US) 0444, 2025 Version 2. The final assessment was June 2025; the syllabus was available only to schools in Arizona, USA. This is not a current exam-entry option. Original self-check teaching and explained practice are not Cambridge papers, official marks or predicted grades. Linked teaching foundations do not establish full syllabus mastery. Calibrated cumulative assessment, reliable drawn-work review and qualified mathematics-teacher review remain unfinished.
Before you begin
- Without-replacement trees
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What should you establish before applying the method in this lesson?
What you will learn
- Read a branch probability after a known first outcome
- Distinguish joint from restricted probability
- Normalise the relevant path probabilities when information is given
Given information changes the question
For 3 red and 2 blue drawn twice without replacement, given that the first was blue, there are 3 red among 4 remaining counters. The chance of red next is 3/4. The joint blue-then-red probability before either draw is (2/5)(3/4)=3/10. These answer different questions.
In a tree, the branch after a known history uses that history as its reference. For an event B given A with P(A)>0, take the probability of both and divide by the probability of the given event. This conditional notation is a useful explanation of tree reasoning; formal set notation is not a prerequisite.
Restrict, then compare
Given at least one red in those two draws, the possible paths are RR,RB,BR, each with probability 3/10. Their total is 9/10. Of that restricted group, RR occupies (3/10)/(9/10)=1/3. Do not keep denominator 1 after ruling out BB. If the given event has probability zero, this division cannot define a conditional probability.
Sources for this lesson
Pause and explain
Is P(second red given first blue) equal to P(first blue and second red)?
Worked example
In the stated unreplaced experiment, find P(second red given first blue).
Show the worked solution
- After blue there are 3 red and 1 blue.
- Use red count over four remaining counters.
Answer 3/4, rather than the joint 3/10.
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.
From 3 red and 2 blue without replacement, find P(second blue given first red).
Give me a hint
Inspect the branch after red.
Compare my reasoning
- Two blue remain among four counters.
- The branch probability is 2/4.
1/2.
Look for these in your work
- Use the known history
- Do not multiply by the first-red probability
Under the same model, find P(both red given at least one red).
Give me a hint
Restrict to RR,RB,BR.
Compare my reasoning
- P(RR)=3/10 and P(at least one red)=9/10.
- Their ratio is 1/3.
1/3.
Look for these in your work
- Remove the excluded BB path
- Normalise by the restricted total
A bag has only one red counter. After two draws without replacement, someone asks for a probability given that both draws were red. Can the ordinary conditional ratio be used?
Give me a hint
The given event cannot occur.
Compare my reasoning
- The probability of two red is zero.
- Dividing by the given-event probability would divide by zero.
No; the ordinary conditional ratio is undefined for this impossible given event.
Look for these in your work
- Check the given event is possible
- Avoid division by zero
Common mistakes
- A conditional branch and a complete path are different probabilities.
- The probability of the given event must be positive.
What can you explain now?
Is P(second red given first blue) equal to P(first blue and second red)?
Compare with the explanation
No; the first uses the known blue history as its reference.
The restricted probability is 3/4, whereas the joint probability is 3/10.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Is P(second red given first blue) equal to P(first blue and second red)?
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.