Projection / capability. Count admissible configurations and convert counts into probabilities only when a probability model is actually specified.
Question this mathematics answers. How many ways can the event occur, and how probable is it under the stated weighting model?
Equation family
Binomial count: C(n,k)=n!/[k!(n-k)!]Binomial probability: P(K=k)=C(n,k)pᵏ(1-p)ⁿ⁻ᵏPascal recursion: C(n,k)=C(n−1,k−1)+C(n−1,k)
Symbols. n = number of independent binary trials/positions; k = number in the designated state; p = model probability of that state on one trial; K = random count.
Assumptions. For the binomial probability, trials are modeled as independent with a common probability p. The combinatorial count itself does not require probability.
Where the inputs come from. n and k come from the operational definition of the event. p should come from calibration data, a validated model, or an explicitly hypothetical assumption—not from moving a slider until the answer looks useful.
Short derivation / justification
Derivation: A specific pattern with k designated outcomes and n−k alternatives has probability pᵏ(1−p)ⁿ⁻ᵏ by independence. There are C(n,k) distinct choices for which k positions carry that outcome. Multiply count × one-pattern weight.
Worked example
Eight independent channels each have p=0.12 false-flag probability. Exactly two flags: C(8,2)=28 compatible patterns; each has weight 0.12²·0.88⁶≈0.00669; total ≈0.1873.
Cadet or real-world application. Fault combinations, packet/relay success counts, reliability screening, experiment outcomes.
What to look for. Changing p changes weights, not the possibility structure. For fixed n,k, P(K=k) is not generally monotonic in p; it peaks near p=k/n.
Common misinterpretation. “More p always means more probability of exactly k events.”
Break / domain boundary. The model says how probability is assigned under its assumptions; it does not identify which channel actually failed or prove independence.