Determining probability of a normal random variable at a single point? I tried to do this exercise from A first course in probability by Sheldon Ross:

An image is partitioned into two regions, one
  white and the other black. A reading taken from
  a randomly chosen point in the white section will
  give a reading that is normally distributed with $μ =
4$ and $σ^2 = 4$, whereas one taken from a randomly chosen point in the black region will have a normally
  distributed reading with parameters $(6, 9)$.
  A point is randomly chosen on the image and has
  a reading of $5$. If the fraction of the image that is
  black is $α$, for what value of α would the probability
  of making an error be the same, regardless of
  whether one concluded that the point was in the
  black region or in the white region?

My attempt at the problem:
Let $A$ be the event that the chosen point has a reading of $5$, and $B$ that it's in the black region.
We need to show that $P(B\mid A)=P(B'\mid A)$, which is equivalent to $P(B\mid A)=\frac12$.
We know that
$$P(B\mid A)=\frac{P(B)P(A\mid B)}{P(B)P(A\mid B)+P(B')P(A\mid B')}$$
Here is my problem. Isn't $P(A)=0$, because of the normal random variable being continuous? If so, how can I compute $P(A\mid B)$ and $P(A\mid B')$? Or maybe my whole approach is wrong?
 A: The normal distribution with expectation $\mu$ and standard deviation $\sigma$ is
$$
\frac 1 {\sqrt{2\pi\,}} \exp\left( \frac{-1} 2 \left( \frac{x-\mu} \sigma \right)^2 \right) \, \frac{dx} \sigma.
$$
Here's what I would do after that:
\begin{align}
& \frac{\Pr(\text{black}\mid \text{data})}{\Pr(\text{white}\mid \text{data})}
= \frac{\Pr(\text{black})}{\Pr(\text{white})} \cdot \frac{\frac 1 {\sigma_1} \exp\left( \frac{-1} 2 \left( \frac{x-\mu_1} {\sigma_1} \right)^2 \right) }{\frac 1 {\sigma_2} \exp\left( \frac{-1} 2 \left( \frac{x-\mu_2} {\sigma_2} \right)^2 \right)} \\[10pt]
= {} & \frac{\Pr(\text{black})}{\Pr(\text{white})} \cdot \frac{\sigma_2}{\sigma_1} \cdot \exp\left( \frac{-1} 2 \left( \left( \frac{x-\mu_1} {\sigma_1} \right)^2 - \left( \frac{x-\mu_2} {\sigma_2} \right)^2 \right) \right) \\[10pt]
= {} & \frac{\Pr(\text{black})}{\Pr(\text{white})} \cdot \frac 3 2 \cdot \exp\left( \frac{-1} 2 \left( \frac 1 9 - \frac 1 4 \right) \right) \\[10pt]
= {} & \frac \alpha {1 - \alpha} \cdot \frac 3 2 \cdot \exp\left( \frac 5 {72} \right).
\end{align}
(Here I used $\mu_1=6$, $\sigma_1=3$, $\mu_2 =4$, $\sigma_2 = 2$, $x=5$.)
Now set that equal to $1$ and solve for $\alpha$.
A: You don't compute the conditional probability mass, you use the conditional probability density.
$$\mathsf P(B\mid A) =\dfrac{\mathsf P(B)f(A\mid B)}{\mathsf P(B)f(A\mid B)+\mathsf P(B')f(A\mid B')}$$
Where $f(A\mid B) = \dfrac{\exp(-(5-4)^2/8)}{\sqrt{8\pi~}}$, $f(A\mid B') = \dfrac{\exp(-(5-6)^2/18)}{\sqrt{18\pi~}}$ 
