Reverse a tree condition using joint and total probabilities
Combine alternative starting sources, then divide the relevant joint path by the total probability of the observed result.
Exam scope and assessment notes
Probability & Statistics 1 (Paper 5, AS & A Level) · Mathematics 9709, 2026–2027, section 5.3. Define the sample space and state its probability assumptions before counting. Fair independent dice give equiprobable ordered pairs; their sums are not equiprobable. Random selection means each specified individual is equally likely under the stated sampling model. Without replacement, later branch probabilities change; with replacement and independent mixing, they remain the same. A′ is a complement, A ∩ B means both, A ∪ B means either or both, and P(A|B) means A given B with P(B) > 0. Mutually exclusive and independent are different properties. The explicit general addition formula is not required by this section; explain overlap through disjoint regions. Conditional probabilities may use P(A|B) = P(A ∩ B)/P(B). Numerical models describe the stipulated experiment, not automatically a real population. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Multiply down probability trees and add disjoint paths.
- Calculate conditional probabilities with a positive denominator.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which paths contribute to total red probability?
What you will learn
- Calculate an observed-result probability across disjoint source paths.
- Recover a source probability given the observed result using a conditional ratio.
Weight each starting source
Choose box A with probability 1/3 or box B with probability 2/3. A contains four red and one blue; B contains one red and four blue. After choosing a box, choose a ball uniformly from it. The joint paths A-and-red and B-and-red have probabilities 4/15 and 2/15, so total red probability is 6/15 = 2/5.
The average (4/5 + 1/5)/2 = 1/2 would wrongly give equal starting weights to the boxes. Starting-source probabilities belong in the tree. Its four joint leaves are 4/15, 1/15, 2/15 and 8/15, summing to one. A red result can arise from either source, so include both source paths in its denominator.
Condition on what was observed
Given a red ball, P(A|red) = P(A and red)/P(red) = (4/15)/(6/15) = 2/3. This differs from P(red|A) = 4/5 and from the prior box-A probability 1/3. Given blue instead, P(B|blue) = (8/15)/(9/15) = 8/9.
The method is a simple conditional ratio; no general symbolic theorem needs to be memorised to solve this numerical tree. In a stipulated item-production model, source weights and mark rates likewise determine the chance of a source given a mark. Such a probability describes the supplied model and should not be claimed as verified real manufacturing evidence.
Pause and explain
The box model has A-and-red 4/15 and total red 6/15. What is P(A|red)?
Worked example
Choose A with probability 1/3, B with 2/3. Red probabilities within A and B are 4/5 and 1/5. Given red, find P(A).
Show the worked solution
- Joint A-and-red is (1/3)(4/5) = 4/15.
- Total red is 4/15 + (2/3)(1/5) = 6/15.
- Divide joint by observed total: (4/15)/(6/15) = 2/3.
Answer P(A|red) = 2/3.
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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.
Choose box A with probability 1/3 or B with 2/3. A has red probability 4/5; B has red probability 1/5. Find total P(red).
Give me a hint
Multiply each source weight by its red branch, then add.
Compare my reasoning
- A-and-red contributes 4/15.
- B-and-red contributes 2/15.
- Their sum is 6/15 = 2/5.
2/5.
Look for these in your work
- Use unequal starting box weights.
- Include both disjoint red paths.
For the same box model, given that the ball is blue, find P(box B | blue).
Give me a hint
Use B-and-blue as numerator and both blue paths in the denominator.
Compare my reasoning
- B-and-blue has probability (2/3)(4/5) = 8/15.
- A-and-blue has probability (1/3)(1/5) = 1/15, so total blue is 9/15.
- The conditional ratio is (8/15)/(9/15) = 8/9.
8/9.
Look for these in your work
- Keep the condition direction.
- Use the total observed-blue probability as denominator.
In a stipulated model, 60% of items come from machine A and 40% from B. A marks 10% of its items; B marks 25%. Given a marked item, find P(it came from B).
Give me a hint
The source-and-mark contributions are 0.60 × 0.10 and 0.40 × 0.25.
Compare my reasoning
- A-and-mark is 0.06; B-and-mark is 0.10.
- Total marked probability is 0.16.
- P(B|marked) = 0.10/0.16 = 5/8.
5/8 in the stipulated model.
Look for these in your work
- Weight each source before conditioning.
- Limit the conclusion to the supplied model.
Common mistakes
- Weight alternative sources by their actual starting probabilities.
- Reverse conditioning by a joint-over-observed-total ratio.
- Do not confuse prior source, source-given-result and result-given-source probabilities.
What can you explain now?
Why does a red observation change box-A probability from 1/3 to 2/3?
Compare with the explanation
A contributes four of the six fifteenths of total red probability.
A red result is more likely from A under this model, so the conditional source weights differ.
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Make it stick
Multiply source paths, add all paths for the observation, then divide the requested joint path by that total.
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.