Zadanie 1c, Paper 3, maj 2025, TZ1, Matematyka AA IB HL
Wzorowane na: Matematyka AA IB HL — maj 2025, TZ1, Paper 3 (zadania wzorowane), pytanie 1(c). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
Darmowe konto pozwoli wrócić do niego później.
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 GM′(t).
(ii) Hence show that GM′(1) = E(M).
Korepetytor
Nie wiesz, jak ruszyć? Korepetytor zna to zadanie i poprowadzi Cię krok po kroku, bez gotowca.
3 rozmowy · bez karty · Zaloguj się
Sprawdź się
Jak poszło?
Pokaż rozwiązanieUkryj rozwiązanie
Odpowiedź z klucza
(i) GM′(t) = (7 + 10t + 9t2 + 4t3)/16. (ii) Evaluating at one gives (7 + 10 + 9 + 4)/16 = 30/16 = 15/8, equal to the expectation in (a).
Schemat punktowania
(i) A2: all four derivative terms correct; A1 for exactly three correct terms. (ii) M1: substitute t=1 into the derivative. A1: correct substituted sum. A1: correct simplification establishing equality. With an incorrect derivative, at most M1 A1 A0 in (ii), even if its value happens to match the expectation.
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.