Zadanie 1f, Paper 3, maj 2024, TZ1, Matematyka AA IB HL
Wzorowane na: Matematyka AA IB HL — maj 2024, TZ1, Paper 3 (zadania wzorowane), pytanie 1(f). 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 production line compares two randomized inspection policies. Under policy A, each batch is inspected independently with constant probability p, where 0<p<1; X is the number of batches up to and including the first inspection. You may use . Under policy B, if the preceding j−1 batches since the last inspection were not inspected, the conditional probability of inspecting batch j is qj = j2/16 for j=1,2,3,4. After an inspection j resets to 1. Y is the number of batches until the first inspection under B.
Polecenie
Explain why Y≤4.
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If the first three batches are not inspected, the conditional inspection probability on batch 4 is 4²/16=1. Thus waiting longer than 4 batches is impossible.
Schemat punktowania
R1: connect certainty at batch 4 to the first inspection waiting time. Accept q₄=4²/16=1 as a sufficient concise explanation, because q_j is explicitly defined in the stem as conditional on the preceding j−1 non-inspections. An unsupported unconditional claim that every fourth batch is inspected is not sufficient.
Komentarz. Paper 3: calculator and AA HL formula booklet allowed. Retain unrounded intermediate values. Give exact answers or three significant figures unless a different precision is requested.