 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ia/paper-1/4f/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Let

Z_{n}

be the size of generation

n

m = 0 \cdot \frac{1}{12} + 1 \cdot \frac{1}{2} + 2 \cdot \frac{1}{3} + 3 \cdot \frac{1}{12} = \frac{17}{12} .

E Z_{n} = (\frac{17}{12})^{n} .

G (s) = \frac{1}{12} + \frac{1}{2} s + \frac{1}{3} s^{2} + \frac{1}{12} s^{3} .

q

is the smallest solution in

[0, 1]

G (q) = q

. Factoring gives

q^{3} + 4 q^{2} - 6 q + 1 = (q - 1) (q^{2} + 5 q - 1) = 0,

q = (29 - 5) /2

. The probability of production continuing forever is

1 - q = \frac{7 - 29}{2} .

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