Five fictional students, A, B, C, D and E, must fill two roles: presenter and recorder. The roles must go to different people, and every allowed ordered assignment is equally likely. List the assignments beginning with A as presenter. There are four: AB, AC, AD and AE. Now ask a different question: select an unordered pair to prepare a poster together. Does AB describe a different pair from BA?
Write both sample spaces for these five students. A sample space lists the elementary outcomes that the random mechanism can produce. An event is a collection of those outcomes. Before dividing a favorable count by a total count, establish that the elementary outcomes are equally likely. Counting alone cannot establish fairness of a real selection procedure.
Predict
Predict the probability that A and B are selected in either order when the two roles are assigned. Then predict the probability that the selected unordered pair is A and B. The denominators will differ; the probabilities need not. Finally, predict the probability of the more restrictive event “A is presenter and B is recorder.” That event must not silently replace “A and B are selected.”
Question
What distinguishes outcomes?
Favorable outcomes
A and B selected for two roles
Presenter and recorder order
AB and BA
Unordered pair is A and B
Membership only
Pair AB
A presents and B records
Presenter and recorder order
AB only
Experiment
Use the live explorer with four, then six people. Predict the total before revealing it. Switch between ordered roles and unordered pairs without changing the number of people. Inspect the complete list: which entries merge when order is removed? The displayed event consistently means selecting A and B in either order, so you can compare the same event across the two sample spaces.
Count the sample space before dividing
Select two different people from A onward. Predict the number of elementary outcomes and the chance that the selected pair is A and B, in either order. Ordered roles and an unordered pair use different sample spaces.
Commit to a prediction, then reveal the table and fixed-scale plot. Change one control and explain the difference.
All settings and traces are temporary. This explorer does not save an answer or award mastery; keep your assumptions and reasoning in the written practice.
Write down one ordered and one unordered calculation for the same setting. Explain why the numerator changes along with the denominator. Then deliberately make a faulty calculation using the ordered numerator and unordered denominator. Describe which sample-space mismatch caused the error. This repair is more informative than memorizing two formulas without knowing when they apply.
Understand
For the five-person worked case, there are five choices for presenter and four remaining choices for recorder. The product rule gives 5 × 4 = 20 ordered outcomes. Each unordered pair appears exactly twice in that list, once in each order, so there are 20/2 = 10 unordered pairs. The probability of selecting A and B is 2/20 = 1/10 in the role experiment. Uniformly selecting one of the ten pairs also gives 1/10. In contrast, requiring the particular assignment AB gives 1/20.
For n distinct people and two different roles, the ordered count is n(n − 1). Removing order divides this count by 2, since every pair has exactly two arrangements. More generally, selecting r different people into r distinct roles gives n(n − 1)⋯(n − r + 1). This is called a permutation count. If only membership matters, every r-person group occurs r! times, where r! means r × (r − 1) × ⋯ × 1. Dividing by that common multiplicity gives a combination count. The division works because the multiplicity is the same for every group.
Replacement changes the mechanism. If two independent uniform draws from A through E allow the same person twice, there are 25 ordered outcomes, including AA. The event “one A and one B” still has AB and BA, so its probability becomes 2/25. If you collapse ordered draws into unordered multisets, repeated pairs such as AA have one preimage, while distinct pairs such as AB have two. Those resulting multisets are not equally likely. A favorable-count divided by number-of-multisets calculation would therefore be wrong.
Apply
Design a probability question about selecting two objects from a collection. State whether replacement is allowed, whether order matters, and what mechanism makes the elementary outcomes equally likely. Give both an enumeration and a multiplication-rule calculation. For example, in the five-person role experiment, P(A then B) = (1/5)(1/4). The second factor is conditional on A already occupying the first role; it is not an independence claim.
A real classroom assignment may exclude conflicts or require one experienced participant. That changes the allowed groups and possibly their probabilities. Keep the counts tied to the actual rule. If someone says “there are ten pairs, so each must have probability one tenth,” identify the missing justification: the procedure must actually select pairs uniformly.
Continue
The next lesson keeps the elementary outcomes but assigns each one a numerical value. Several different outcomes may produce the same number. Carry forward the distinction between an outcome and its label: just as AB and BA can merge into one pair, different random outcomes can merge into one value without making all values equally likely.
Apply · explain · revise
Practice and reasoning
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Six named learners are eligible for distinct chair and recorder roles, with no person taking both. How many ordered assignments are possible?
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
Among six learners, two are experienced. Uniformly assign chair and recorder without replacement. What is the probability both roles go to experienced learners?
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
Label six learners A–F. Compare a uniform two-person committee with uniform chair/recorder selection. Give sample-space counts, the probability the selected people are exactly A,B, and a small diagram/table explaining the factor of two.
Structure your argument and preserve revisions
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A box has four distinct red tokens and three distinct blue tokens. Draw three without replacement. Find the probability of exactly two red tokens using combinations, then explain an equivalent ordered approach.
Structure your argument and preserve revisions
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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Two names are drawn independently with replacement from five equally likely names. Find the probability the names differ, and explain why 5·4 cannot be the full denominator.
Structure your argument and preserve revisions
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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
A fair coin chooses team A or team B, then a person is drawn uniformly from that team. Team A has two people and B has six. Is the resulting person uniform among all eight? Design a uniform alternative.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Reflect and continue
Before marking this lesson complete, explain one key idea without the worked solution and identify a condition that matters. Marking complete records participation, not independent mastery.
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