Zadanie 1e, Paper 3, maj 2025, TZ1, Matematyka AA IB HL
Wzorowane na: Matematyka AA IB HL — maj 2025, TZ1, Paper 3 (zadania wzorowane), pytanie 1(e). 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
(i) Find p.
(ii) Find GY(t).
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(i) The t-squared coefficient is the probability of two detections: 0.4p = 0.24, so p = 0.6. (ii) P(Y = 0) = 0.6(0.4) = 0.24. Exactly one detection has probability 0.4(0.4) + 0.6(0.6) = 0.52. Thus GY(t) = 0.24 + 0.52t + 0.24t2.
Schemat punktowania
(i) A1: equation using independent detections. A1: p=0.6. (ii) M1: form the distribution polynomial with the value of p. A1: zero-detection probability. A1: one-detection probability. A1: complete correct polynomial. An incorrect p can earn at most three marks in (ii) through consistent working; no final accuracy mark.
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.