Zadanie 1a, Paper 3, maj 2024, TZ1, Matematyka AA IB HL
Wzorowane na: Matematyka AA IB HL — maj 2024, TZ1, 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
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
For policy A with p=.12, find (i) the probability that the first inspection is on batch 4; (ii) the probability of at least one inspection among the first 8 batches.
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(i) Three non-inspections followed by an inspection have probability (.88)³(.12)=.08177664≈.0818. (ii) The complement of no inspections has probability 1−(.88)⁸=.6403654752…≈.640. Equivalently useBin(8,.12) or sum first-inspection probabilities for indices 1..8.
Schemat punktowania
(i) M1:form the ordered product of non-inspection and inspection probabilities. A1:.0818. (ii) M1:use the complement, binomial model or sum of first-inspection paths. A1:correct parameters/expression 1−.88⁸. A1:.640. For (i), source M1 is earned by evidence of a non-inspection probability .88 as part of multiplication, even if the complete path is not yet correct. For (ii), a sum with at least three correct first-inspection terms earns M1. Correct Bin(8,.12) parameters with the erroneous event of more than one inspection give M1A1A0 (probability .2480369027…).
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.