Zadanie 9g, Paper 1, maj 2024, TZ1, Matematyka AA IB SL

The first-violet probabilities for n=1,2,3,4 are 2/6=1/3, (4/6)(3/7)=2/7, (4/6)(4/7)(4/8)=4/21, and (4/6)(4/7)(4/8)(5/9)=20/189. Thus n=4. No later index gives the same value: if pₙ is the first-violet probability, pₙ₊₁/pₙ=4(n+2)/[(n+1)(n+6)]<1 for n≥1, so it strictly decreases. An equivalent product is $p_n=[\prod_{j=0}^{n-2}4/(6+j)](n+1)/(n+5)$.

M1: form successive first-violet paths or a general product, not a single-step conditional probability. A1:correct initial amber-run product. A1:correct final violet probability at the tested index. A1:their product equals 20/189, or correct consecutive probabilities demonstrating the pattern. A1:n=4. A table of at least three successive cases including the matching fourth case is an equivalent full-credit route. As in the source, a correct bare n=4 with no working earns M1A0A0A0A1, 2/5.