There are two ways in which you could try to get the possible outcome for the untested dose (say 0.7).
Preferable way
Since you have covariates for each patient (i.e., age, gender, etc., say $\mathbf x$ (a vector)), dosage ($d$), and the response ($0/1, y$), fitting a logistic regression with $y$ as response and $(d, \mathbf x)$ as covariates is the easiest way to model the dependence of $y$ on $d$. Once you have trained your logistic regression model on the available data, you could predict for any patient in the future, given you have their covariate $(d_{\text{new}}, \mathbf x_{\text{new}})$ information by
$$
p_{\text{new}}=\frac{\exp(\beta_0 + \beta_d d_{\text{new}} + \mathbf{\beta}_{\mathbf x}^' \mathbf x_{\text{new}})}{1 + \exp(\beta_0 + \beta_d d_{\text{new}} + \mathbf{\beta}_{\mathbf x}^' \mathbf x_{\text{new}})}
$$
Because you have trained your logistic regression before, you should have a cutoff $c$ for deciding the possible outcome such that
$$
p_{\text{new}} > C \implies 1
$$
$$
p_{\text{new}} \le C \implies 0
$$
Your way
Lets say you train 3 separate logistic regression models based on 0.1, 0.5, and 0.9 with unequal group size, and you could predict probabilities $p_{0.1}, p_{0.5},$ and $p_{0.9}$ for a particular dose (say 0.7).
A meaningful way of interpolation to determine the probability (only if you insist on it), could be
$$
p_{0.7} = \frac {\displaystyle \sum_{i \in \{0.1, 0.5, 0.9\}} w_i p_i}{\displaystyle\sum_{i \in \{0.1, 0.5, 0.9\}} w_i}
$$
where the weights could be decided in various ways. A few options are:
$w_i = \frac{n_i}{(i-0.7)^2}$, where $i$ is 0.1, 0.5, and 0.9, and $(i-0.7)^2$ reflects the "closeness" of the new dosage to the original 3 groups and $n_i$ is the number of individuals in the ith groups controlling for higher confidence in the weights with higher $n_i$.
$w_i = \frac{n_i}{|i-0.7|}$, where $i$ is 0.1, 0.5, and 0.9, and $|i-0.7|$ reflects the "closeness" of the new dosage to the original 3 groups and $n_i$ is the number of individuals in the ith groups controlling for higher confidence in the weights with higher $n_i$.
However, I would not be very confident in these methods as they are based on a lot of assumptions and "hand-waving".
