Review: Statistics: Counting and Random-Variable Models
Counting methods distinguish ordered roles from unordered groups. A probability model assigns chances to possible outcomes, and expected value or binomial counting combines those probabilities while preserving what a single trial can actually produce.
Bring the unit skills together
Use these 4 new cases to check your reasoning. Choices are temporary and are not saved to an account, sent to staff or recorded as school grades. Download your current review choices before leaving or reloading.
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Use the skills in a new task
Create a four-part mathematical solution sheet for these supplied cases. 1. From 12 distinct candidates choose a chair and a different recorder. How many assignments are possible? Candidate answers: (A) 131 (B) 132 (C) 133 (D) 264. 2. From 12 distinct candidates choose an unordered two-person committee. How many committees are possible? Candidate answers: (A) 132 (B) 65 (C) 66 (D) 67. 3. A supplied fictional payoff model gives +36 points with probability 1/4 and −9 points with probability 3/4. Find expected payoff per trial. Candidate answers: (A) 1.25 points (B) 3.25 points (C) 4.5 points (D) 9/4 points. 4. A fair coin is tossed independently 10 times. Find the probability of exactly one head. Candidate answers: (A) 1/1024 (B) 1/2 (C) 1023/1024 (D) 5/512.
Work on paper or in your own document. Explain the choices you made.
- State each requested quantity or relationship and its input restrictions.
- Show intermediate calculations or the applicable theorem with matching correspondences.
- Give all four results, including exact values or requested rounding and units.
- Check a result by substitution, reversal, dimensions, a representation or a counterexample as appropriate.
Compare with a model response
1. 132. There are 12 choices for chair, then 11 for recorder. Ordered roles distinguish assignments, so multiply to get 132. 2. 66. Ordered selections total 12(11), but each pair is counted twice in reverse order. Divide by 2: 66 committees. 3. 9/4 points. Weight each outcome by its probability: 36/4−3(9)/4=9/4=9/4. Expected value is a long-run mean of this model, not a guaranteed single-trial payoff. 4. 5/512. The single head can occupy any of 10 positions. Each resulting complete sequence has probability (1/2)^10; adding these disjoint sequences gives 5/512. Exactly one excludes all other head counts.
A different response can also meet the criteria. Compare the evidence and reasoning, rather than copying the wording.