Conditional probability of a binary signal conditioned on a continuous observation Let K a Bernoulli Random Variable with parameter 0.1 (p(K=1) = 0.1).
Let X = K + W where W is a noise, a standard Normal variable.
We want to find p(K = 0 | X = 0.5).  I reason as follows:  
Using the Bayes rule:

In the conditional Universe of K = k, X is the sum of a standard Normal variable and a constant k.  Therefore it is a Normal variable N(k, 1). Therefore we have:

Do you agree?
 A: Your calculations are correct. 
Since the conditional densities of $X$ given $K=k$ are $N(k,1)$ $k = 0,1$, they have the same numerical value $\alpha$ at $x = 0.5$, and so $f_X(x)$, the unconditional distribution of $X$, being the weighted sum $0.9f_{X\mid K=0}(x\mid K=0)+ 0.1f_{X\mid K=1}(x\mid K=1)$, also has value $0.9\alpha + 0.1\alpha = \alpha$ at $x=0.5$. Hence, the a posteriori probability of the event $\{K=0\}$ when it is observed that $X$ had value $0.5$ is the same as the a priori probability of the event $\{K=0\}$. For all other possible observed values of $X$, $f_{X\mid K=0}(x\mid K=0) \neq f_{X\mid K=1}(x\mid K=1)$, and the a posteriori probability of the event $\{K=0\}$ is different from the a priori probability of the event $\{K=0\}$.
The a posteriori probability of the event $\{K=0\}$ is a continuous decreasing function of $x, x\in \mathbb R$, where $x$ is the observed value of $X$, decreasing from a limiting value of $1$ at $-\infty$ to a limiting value of $0$ as $x \to \infty$. Thus, there must some $x$ for which this a posteriori probability equals the a priori probability, and in this problem, this equality occurs at $x=0.5$ regardless of the choice of value for the a priori probability in $(0,1)$.
