Determining probability of a normal random variable at a single point?

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 $\alpha$.


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 $f(A\mid B) = \dfrac{\exp(-(5-4)^2/8)}{\sqrt{8\pi~}}$, $f(A\mid B') = \dfrac{\exp(-(5-6)^2/18)}{\sqrt{18\pi~}}$