Counting and combined probability
Decide whether order matters and whether events are independent.
Exam scope and assessment notes
Probability & Statistics 1 (Paper 5) · Mathematics 9709, 2026–2027. This revision overview focuses on selections, assigned roles and event classification; it does not cover every arrangement restriction or all of sections 5.2–5.3. Written solutions use self-check criteria, not an official mark scheme. Specialist accuracy review has not been completed. The diagram uses three names as a separate illustration; use the numbers and random models specified in each task.
Before you begin
- Use factorials and simplify fractions.
- Interpret intersection as both events and recognise equally likely outcomes.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Five candidates are available. Compare choosing two equal-status representatives with appointing a lead and a deputy, with nobody holding both roles. How many outcomes are there?
What you will learn
- Use combinations for unordered selections.
- Distinguish independent from mutually exclusive events.
Count outcomes consistently
A committee is a selection, so rearranging its members does not create a new committee. Use combinations to count selections. In contrast, choosing a first, second and third place distinguishes order and needs a different count.
A probability can be calculated as favourable outcomes divided by all outcomes only when the counted outcomes are equally likely and use a consistent counting method.
Independence is a condition
Independent events satisfy P(A∩B)=P(A)P(B). Mutually exclusive events cannot happen together, so their intersection has probability zero. Events with positive probabilities cannot be both independent and mutually exclusive.
For any two events, P(A∪B)=P(A)+P(B)−P(A∩B). Subtract the intersection so outcomes in both events are not counted twice.
Sources for this lesson
Pause and explain
Events A and B have P(A) = 2/5, P(B) = 1/4 and P(A ∩ B) = 1/20. How are they related?
Worked example
A team of 2 is selected from 7 students. How many teams are possible?
Show the worked solution
- Order does not matter: choosing Ana then Ben is the same team as Ben then Ana.
- There are 7×6 ordered selections.
- Divide by 2! to remove duplicate orderings.
Answer 21 teams
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.
Eight candidates form a four-person panel with one chair, one secretary and two equal-status ordinary members. Nobody holds two roles. Count the possible panels with roles, and check your answer by choosing the chair and secretary first.
Give me a hint
A panel membership set and an assignment of roles are different outcomes. Only the two ordinary members are interchangeable.
Compare my reasoning
- First choose four people in 8C4 = 70 ways. Choose the chair in 4 ways and the secretary from the remaining three in 3 ways. The other two need no ordering.
- The total is 70×4×3 = 840. Dividing all four roles by 4! would erase distinct chair/secretary assignments; multiplying the ordinary pair by 2 would count it twice.
- Alternatively choose the chair in 8 ways, the secretary in 7 ways and the two ordinary members from the remaining six in 6C2 = 15 ways. This gives 8×7×15 = 840.
840 panels with roles.
Look for these in your work
- Separate membership selection from role assignment.
- Account for exactly which positions are interchangeable.
- Obtain the same total by two valid counting orders.
Choose an unordered set of three tickets uniformly from eight distinct tickets numbered 1–8. Let A mean ticket 1 is included and B mean ticket 2 is included. Find P(A), P(B) and P(A ∩ B). Decide whether A and B are independent or mutually exclusive, and check using P(B | A).
Give me a hint
Count all three-ticket sets once. For the intersection, both named tickets are already fixed.
Compare my reasoning
- There are 8C3 = 56 equally likely sets. Sets containing ticket 1 number 7C2 = 21, so P(A) = 21/56 = 3/8; symmetry gives P(B) = 3/8.
- Sets containing both named tickets number 6C1 = 6. Thus P(A ∩ B) = 6/56 = 3/28. This is positive, so the events are not mutually exclusive. It differs from (3/8)² = 9/64, so they are not independent.
- Given A, the other two tickets are chosen uniformly from seven, so P(B | A) = 2/7. Equivalently 6/21 = 2/7. This is below 3/8: fixing one place for ticket 1 reduces the chance that ticket 2 is also included.
P(A) = P(B) = 3/8; P(A ∩ B) = 3/28; P(B | A) = 2/7. Neither independent nor mutually exclusive.
Look for these in your work
- Use a consistent unordered sample space and state its size.
- Compare intersection with both zero and the product of marginals.
- Restrict the denominator correctly when conditioning on A.
A token is chosen uniformly from 12 distinct tokens numbered 1–12. A means its number is divisible by 3; B means divisible by 4; C means the number is 1 or 2. A learner claims that overlapping events must be dependent, whereas mutually exclusive events must be independent. Evaluate both claims by comparing A with B, then A with C.
Give me a hint
List event members before calculating. An example can refute a claim about all events.
Compare my reasoning
- A = {3, 6, 9, 12} and B = {4, 8, 12}. Their probabilities are 1/3 and 1/4, and their intersection is {12} with probability 1/12. This equals (1/3)(1/4), so A and B are independent despite overlapping.
- C = {1, 2} has probability 1/6. A and C have no common outcome, so they are mutually exclusive. But 0 differs from (1/3)(1/6) = 1/18, so they are not independent.
- Both claims fail. Overlap asks whether both events can occur; independence asks whether the joint probability equals the product. Two mutually exclusive events with positive probabilities cannot be independent.
A and B: independent, not mutually exclusive. A and C: mutually exclusive, not independent.
Look for these in your work
- Enumerate event members and intersections correctly.
- Test independence numerically rather than inferring it from overlap.
- Explain why positive marginal probabilities matter in the mutual-exclusion argument.
Common mistakes
- Do not multiply event probabilities unless independence or the appropriate conditional probability justifies it.
- A committee is not an ordered arrangement.
What can you explain now?
Independent events have probabilities 0.3 and 0.5. Find the probability that both occur.
Compare with the explanation
0.15
Independence permits multiplication: 0.3×0.5=0.15.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Explain when swapping selected people creates a new outcome. Then distinguish testing an intersection against zero from testing it against the product of the marginal probabilities. State how a conditional sample space changes the denominator.
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.