I was preparing for an exam in Image Analysis, when I found this probability problem in one of the old exam sets.
A bunch of products is measured, and two features $(x_1,x_2)$ are measured on each. A product is either fine, or defect. If the product is fine, it is assumed that its features belongs to a bivariate normal distribution
$\mu = \begin{pmatrix} 0.5\\0.5 \end{pmatrix} , \;\; \Sigma = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} $
If the product is defect, it belongs to a uniform distribution
$ f\left( \begin{pmatrix} x_1\\ x_2 \end{pmatrix} |\text{ defect}\right) = 1, x_1 \in [0;1], x_2 \in[0;1] $
The prior probability of being either defect or fine, is equal.
Now we are given a product with features $(0.2, 0.3)$, and we want to determine with what probability the product was fine.
I tried to find the probability of being the region $x_1\in[0;1], x_2 \in [0,1]$ given that the product was fine, and found (using a double integral of the density function of the bivariate normal distribution over the region) that this was approx. 0.6. Then I tried applying Bayes rule, but this did not work out well.