One solution to this problem is the forward-backward algorithm, which produces the probabilities that you need. The details of the computation are for example described in Section III.B of Rabiner (1989), A Tutorial on Hidden Markov Models, and I include some of the equations below (trying to stay as close as possible to Rabiner's notation).
Notation
Following Rabiner (1989), consider the notation:
- $O_t$ = observation at time $t$ (e.g., $O_t \in \{ \text{Umbrella}, \text{No Umbrella}\}$)
- $q_t$ = state at time $t$; for simplicity, the states can be indexed by positive integers (e.g., $q_t \in \{ 1, 2\}$ if there are two possible states)
- $\lambda$ = vector of all model parameters (initial distribution, transition probabilities, and state-dependent observation parameters)
- $T$ = length of the time series, i.e., time goes from $t = 1$ to $t = T$
- $N$ = number of states (e.g., $N=2$ in your example for rain and no rain)
Forward and backward variables
The forward variable $\alpha_t(i)$ is defined (Rabiner, Equation 18) as
$$
\alpha_t(i) = \Pr(O_1, O_2, \dots, O_t, q_t = i \mid \lambda),
$$
i.e., the probability (or probability density for continuous observations) of observing the sequence $O_1, \dots, O_t$ up to time $t$ and being in state $i$ at time $t$, given the model described by $\lambda$.
Similarly, the backward variable $\beta_t(i)$ is defined (Rabiner, Equation 23) as
$$
\beta_t(i) = \Pr(O_{t+1}, O_{t + 2}, \dots, O_T \mid q_t = i, \lambda),
$$
i.e, the probability/density of observing a given sequence between times $t+1$ and $T$, given that the state at time $t$ is $i$, and given the model parameters $\lambda$.
Both forward and backward variables can be computed relatively easily using iterative algorithms, sometimes called the "forward" and "backward" algorithms, described in Equations 19-20 and 24-25 of Rabiner (1989), respectively.
State probabilities
The quantity that you are interested in is the state probability
$$
\gamma_{t}(i) = \Pr(q_t = i \mid O_1, \dots, O_T, \lambda),
$$
for all times $t \in \{ 1, \dots, T\}$ and states $i \in \{ 1, \dots, N\}$. From the definition of conditional probability,
$$
\gamma_{t}(i) = \frac{\Pr(q_t = i, O_1, \dots, O_T \mid \lambda)}{\Pr(O_1, \dots, O_T \mid \lambda)}
$$
where
$$
\begin{aligned}
\Pr(q_t = i, O_1, \dots, O_T \mid \lambda) & = \Pr(O_1, \dots, O_t, q_t = i \mid \lambda) \times \Pr(O_{t+1}, \dots, O_T \mid q_t = i, \lambda) \\
& = \alpha_t(i) \beta_t(i)
\end{aligned}
$$
From the law of total probability, we can also write the denominator as
$$
\begin{aligned}
\Pr(O_1, \dots, O_T \mid \lambda) & = \sum_{k=1}^N \Pr(q_t = k, O_1, \dots, O_T \mid \lambda) \\
& = \sum_{k=1}^N \alpha_t(k) \beta_t(k)
\end{aligned}
$$
Finally, the state probabilities can be computed from the forward and backward variables as
$$
\gamma_t(i) = \frac{\alpha_t(i)\beta_t(i)}{\sum_{k=1}^N \alpha_t(k)\beta_t(k)}
$$
In conclusion, to compute the state probabilities, we would start by running the forward and backward algorithms (collectively known as the forward-backward algorithm), and then use the last formula above.
