Zadanie 1a, Paper 3, maj 2026, TZ2, Matematyka AA IB HL
Wzorowane na: Matematyka AA IB HL — maj 2026, TZ2, Paper 3 (zadania wzorowane), pytanie 1(a). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
Wspólne informacje do wiązki zadań
An energy-buffer simulation has integer charge levels 0,1,…,w. While the charge is strictly between 0 and w, each independent tick increases it by one with probability p and decreases it by one with probability q=1−p, where 0<p<1 and p≠q. The simulation stops as soon as charge reaches 0 (shutdown) or w (full charge). Let Rₙ be the probability of reaching w before 0, starting at charge n. Thus R₀=0, R_w=1, and Rn = qRn−1 + pRn+1 for 1 ≤ n ≤ w − 1.
Polecenie
Initially w=3, p=1/3 and charge is 1. (i) Show that the probability a of still running with charge 1 after exactly two ticks is 2/9. (ii) Find the probability b that full charge is reached for the first time on tick 4.
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Odpowiedź z klucza
(i) The only permitted two-tick return is up then down, probability (1/3)(2/3)=2/9; down first ends the run. (ii) First reaching 3 on tick 4 requires up,down,up,up. Thus b=(1/3)(2/3)(1/3)(1/3)=2/81≈0.0247. Reaching full on tick 2 is not part of this event.
Schemat punktowania
(i) A1: product (1/3)(2/3), establishing 2/9. (ii) M1: multiply their a by the two upward probabilities, or use the full four-tick path. A1: 2/81 or 0.0247, with consistent FT.
Komentarz. Paper 3: calculator and AA HL formula booklet allowed. Give exact values or three significant figures unless requested otherwise; proofs and numerical solutions require supporting work.