Use selections and ordered assignments to calculate probability
Apply counting methods to favourable and total outcomes using the same sampling model.
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
- Calculate permutations and combinations.
- Distinguish membership from named roles.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which count describes unordered three-ball samples from eight distinct balls?
What you will learn
- Use combinations for equally likely unordered samples.
- Use permutations for ordered assignments without mixing outcome conventions.
A random subset uses combinations
A bag has four distinct blue and three distinct red balls. Choose three without replacement, with every three-ball subset equally likely. There are 7C3 = 35 total subsets. Exactly two blue and one red gives 4C2 × 3C1 = 18 favourable subsets, so probability 18/35. Same-colour balls remain distinct physical individuals when forming the sample space.
For five red and three blue balls, a random three-ball subset has 8C3 = 56 possibilities. Exactly two red and one blue gives 5C2 × 3C1 = 30, hence 30/56 = 15/28. Do not treat colour-composition labels as equally likely outcomes; they can represent very different numbers of subsets.
An assignment uses ordered outcomes
Six equally likely no-tie competitors fill first, second and third places. The model has 6P3 = 120 equally likely podiums. If A and B must both appear, choose the third competitor from the other four, then arrange the three in 3! ways: 24 favourable podiums. Their probability is 24/120 = 1/5.
A matching unordered method chooses three podium members: favourable sets 4 and all sets 6C3 = 20, also giving 1/5, because all six orders within each set have the same probability in this model. Dividing an unordered favourable count by an ordered total would be inconsistent. Equal podium probabilities are a stipulated mathematical model, not a claim about competitors’ real abilities.
Pause and explain
Five red and three blue: what is P(exactly two red in a random three-ball subset)?
Worked example
Choose three balls at random without replacement from five distinct red and three distinct blue balls. Find P(exactly two red).
Show the worked solution
- All unordered samples number 8C3 = 56.
- Choose two red and one blue in 5C2 × 3C1 = 30 ways.
- The probability is 30/56 = 15/28 under uniform subset sampling.
Answer 15/28.
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.
Select two books uniformly without replacement from two distinct fiction and three distinct non-fiction books. Find P(both fiction).
Give me a hint
There is one favourable fiction pair among all two-book subsets.
Compare my reasoning
- The total subset count is 5C2 = 10.
- Both fiction gives 2C2 = 1 favourable pair.
- Divide 1 by 10 to get 1/10.
1/10.
Look for these in your work
- Use a uniform unordered-pair model.
- Keep favourable and total counts consistent.
Select three balls uniformly without replacement from five distinct red and three distinct blue balls. Find P(exactly two red).
Give me a hint
Choose two red individuals and one blue individual.
Compare my reasoning
- All subsets number 8C3 = 56.
- Favourable subsets number 5C2 × 3C1 = 30.
- The probability is 30/56 = 15/28.
15/28.
Look for these in your work
- Apply the exact composition.
- Simplify the ratio after counting.
In a stipulated model with 120 equally likely no-tie podiums from six competitors A–F, find P(A and B both appear). Check with an unordered method.
Give me a hint
Choose the third member, then assign the three places.
Compare my reasoning
- Four third-member choices and six place orders give 24 favourable podiums.
- Their probability is 24/120 = 1/5.
- Unordered membership gives 4/6C3 = 4/20 = 1/5 as well.
1/5 by both consistent methods.
Look for these in your work
- Retain both required people.
- Use a consistent outcome convention in each ratio.
Common mistakes
- Use the same ordered or unordered convention in both counts.
- Treat each physical ball as an individual in the sample space.
- Equal likelihood belongs to the stated model, not a real-world ranking assumption.
What can you explain now?
Can an unordered favourable count be divided by an ordered total?
Compare with the explanation
No; numerator and denominator must count the same kind of outcome.
Otherwise one count includes extra orderings that the other omits.
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Make it stick
Count favourable and total outcomes under one sampling model; use sets for membership and ordered assignments for roles.
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.