Solve unknown probability parameters and mixed conditions
Use model equations, restrict the conditional universe and reject parameter values outside the physical assumptions.
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
- Solve a quadratic and check admissible roots.
- Use combinations, conditional ratios and probability trees.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which value can represent a number of red balls?
What you will learn
- Infer an unknown count or source weight from a supplied probability.
- Combine selection restrictions with conditioning and verify the result.
An unknown count must remain a valid integer
A bag has r red and two blue balls. Draw two uniformly without replacement; suppose P(both red) = 2/5. Then r(r − 1)/[(r + 2)(r + 1)] = 2/5. Rearranging gives 3r² − 11r − 4 = 0 = (3r + 1)(r − 4), so r = 4 or r = −1/3. Only the non-negative integer count r = 4 fits the bag model.
Substitute back: (4/6)(3/5) = 2/5. A probability equation can have algebraic solutions that violate count, positivity or denominator conditions. Always verify the chosen value in the original experiment rather than relying only on a rearranged equation.
Conditioning and selection restrictions change different parts
Choose three people uniformly from five A-group and three B-group candidates in disjoint groups. Given at least one B was selected, all-A teams are removed, leaving 8C3 − 5C3 = 46 equally likely eligible subsets. Exactly two B has 3C2 × 5C1 = 15 favourable subsets, so the conditional probability is 15/46, not the unconditional 15/56.
Choose box A with unknown probability q and B with 1 − q; their red probabilities are 3/4 and 1/4. If total red probability is 1/2, then (3/4)q + (1/4)(1 − q) = 1/2, giving q = 1/2. Given red, the A probability is [(1/2)(3/4)]/(1/2) = 3/4. The source weight and reverse conditional are distinct quantities, both within [0,1].
Pause and explain
Five A and three B candidates; choose three, given at least one B. What is P(exactly two B)?
Worked example
A bag has r red and two blue; two draws without replacement have P(both red) = 2/5. Find r.
Show the worked solution
- Form r(r − 1)/[(r + 2)(r + 1)] = 2/5.
- Rearrange to (3r + 1)(r − 4) = 0, giving 4 or −1/3.
- Keep integer r = 4 and check (4/6)(3/5) = 2/5.
Answer r = 4.
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.
A bag has r red and two blue. Two uniform draws without replacement have P(both red) = 2/5. Find r and verify it.
Give me a hint
Write the sequential fraction, then reject roots that cannot be individual counts.
Compare my reasoning
- Set r(r − 1)/[(r + 2)(r + 1)] = 2/5.
- Rearrange to (3r + 1)(r − 4) = 0; valid count r = 4.
- Check (4/6)(3/5) = 2/5 in the original model.
r = 4; the root −1/3 is inadmissible.
Look for these in your work
- Show the probability equation.
- Check integer and original-denominator conditions.
Choose three uniformly from five A and three B candidates in disjoint groups. Given at least one B, find P(exactly two B).
Give me a hint
Remove all-A teams from the conditional total.
Compare my reasoning
- All subsets number 56; all-A subsets number 10, leaving 46 eligible teams.
- Exactly two B gives 3C2 × 5C1 = 15 subsets.
- The conditional probability is 15/46.
15/46.
Look for these in your work
- Apply the given information to the denominator.
- Count exactly two B without ordering team members.
Choose box A with probability q or B with 1 − q. Red probabilities are 3/4 in A and 1/4 in B. Total P(red) is 1/2. Find q, then P(A|red).
Give me a hint
First solve the weighted total, then use a joint-over-observed-total ratio.
Compare my reasoning
- (3/4)q + (1/4)(1 − q) = 1/2 gives q = 1/2.
- A-and-red is (1/2)(3/4) = 3/8.
- P(A|red) = (3/8)/(1/2) = 3/4; both values meet the probability bounds.
q = 1/2; P(A|red) = 3/4.
Look for these in your work
- Separate the unknown source weight from the reverse condition.
- Verify the total and admissible probability range.
Common mistakes
- Reject non-integer or negative ball counts.
- Use a conditional denominator after excluding incompatible subsets.
- Check source weights and probabilities against their [0,1] domains.
What can you explain now?
Why is 15/56 not the conditional team probability here?
Compare with the explanation
It includes ten all-A teams that contradict the given at-least-one-B information.
The conditioned total is 46; the favourable fifteen teams remain eligible.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Build the model equation, restrict the conditional universe, and verify every inferred parameter in the original experiment.
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.