Hidden Markov Model Probability

Question

I'm trying to understand how to find the probability of a given HMM.

Here's the question that I have: Find $p(acca|EE5I)$

I did: $$\begin{align}p(acca|EE5I) &= E(a)*E(c)*5(c)*I(a)\\ &= .25 * .25 *.95 * .40\\ &= .02375 \end{align}$$ However, when I'm not given the node $EE51$, I'm confused as to how to compute this probability:

$p(acca)$

Answer 1

First note that to have $acca$, either $E$ or $I$ must be repeated. There are four different ways to do that, and for each we can find its probability:

\begin{align} P(EEEE) &= 0.9^3 \\ P(EEE5) &= 0.9^2 \cdot 0.1 \\ P(EE5I) &= 0.9 \cdot 0.1 \\ P(E5II) &= 0.9 \cdot 0.1\end{align}

These probabilities add to 0.99 because with probability 0.01 we observe $E5I\text{End}$. We can then compute the probabilities of $acca$ conditional to each of these paths:

\begin{align} P(acca \mid EEEE) &= 0.25^4 \\ P(acca \mid EEE5) &= 0.25^3 \cdot 0.05 \\ P(acca \mid EE5I) &= 0.25^2 \cdot 0.95 \cdot 0.4 \\ P(acca \mid E5II) &= 0.25 \cdot 0.95 \cdot 0.1 \cdot 0.4\end{align}

Finally,

\begin{align} P(acca) &= P(acca \mid EEEE) \cdot P(EEEE) + \\ &\quad P(acca \mid EEE5) \cdot P(EEE5) + \\ &\quad P(acca \mid EE5I) \cdot P(EE5I) + \\ &\quad P(acca \mid E5II) \cdot P(E5II) \\ &= 0.0059034 \end{align} barring mistakes.