Zadanie 1d, Paper 3, maj 2025, TZ1, Matematyka AA IB HL
Wzorowane na: Matematyka AA IB HL — maj 2025, TZ1, Paper 3 (zadania wzorowane), pytanie 1(d). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
Wspólne informacje do wiązki zadań
A polynomial can encode a finite probability distribution: for a nonnegative integer-valued random variable U, define . Two independent random selectors each return one of 1, 2, 3, 4 with equal probability. Let M be the smaller of the two outputs. Its probabilities for m = 1, 2, 3, 4 are respectively 7/16, 5/16, 3/16, 1/16, so GM(t) = (7t + 5t2 + 3t3 + t4)/16. In later parts, X counts experts in a two-person panel chosen uniformly from seven people, three experts and four trainees. Independently of that selection, two sensors attempt to register a pulse; their independent detection probabilities are 0.4 and p. Let Y count detections. The coefficient of t2 in GY(t) is 0.24. For Z = X + Y, you may use GZ(t) = GX(t)GY(t).
Polecenie
Show that , explaining the probability calculation for every coefficient.
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There are equally likely two-person panels. No experts: . Exactly one expert: . Two experts: . Multiplying each probability by its corresponding power of t gives the stated polynomial.
Schemat punktowania
M1: attempt a valid without-replacement probability calculation. A1: worked probability for zero experts. A2: worked probability for exactly one expert; if using ordered draws, award one accuracy mark for each order. A1: worked probability for two experts, also obtainable by complement. Merely repeating the given coefficients earns no accuracy marks.
Komentarz. Paper 3: calculator use is permitted, but reasoning must be shown. Exact equivalent forms and appropriate numerical answers to three significant figures are accepted unless stated otherwise.