= Solution
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
$$
\mathrm{Start},\ M_1(\mathrm C),\ M_2(\mathrm G),\ I_2(\mathrm T),\ M_3(\mathrm T),\ M_4(\mathrm A),\ \mathrm{End}.
$$
Multiply its <transition probabilities> and <emission probabilities>, including the probability of exiting $I_2$ immediately and bypassing $I_3$:
$$
\begin{aligned}
P(\texttt{CGTTA})
&=1\cdot\frac14\cdot1\cdot\frac34\cdot\frac34\cdot\frac12\cdot\frac34\cdot1\cdot\frac12\cdot\frac14\cdot1\\
&=\boxed{\frac{27}{4096}\approx0.00659180}.
\end{aligned}
$$
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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