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ID: past-exam-of-the-mathematical-and-computational-biology-course-of-the-university-of-cambridge/2025/paper-2/2/b/solution

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Past exam of the mathematical and Computational Biology course of the University of Cambridge / 2025 / Paper 2 / 2 / b / Solution by

Codex  0 2026-10-20

The sequence has five characters, while the four match states account for four compulsory emissions. Consequently exactly one insertion state emission is needed. An insertion at

I_{3}

would have to emit T, which has probability zero there. The only nonzero hidden path is therefore

Start, M_{1} (C), M_{2} (G), I_{2} (T), M_{3} (T), M_{4} (A), End .

(1)

Multiply its transition probabilities and emission probabilities, including the probability of exiting

I_{2}

immediately and bypassing

I_{3}

P (CGTTA) = 1 \cdot \frac{1}{4} \cdot 1 \cdot \frac{3}{4} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{3}{4} \cdot 1 \cdot \frac{1}{2} \cdot \frac{1}{4} \cdot 1 = \frac{27}{4096} \approx 0.00659180 .

(2)

Normally the forward algorithm sums over all compatible paths. Here all other paths have probability zero, so this single path gives the full sequence probability.

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