, 2020

Solution:

Under the transformation $(x, y) \mapsto (3x+2y, x+4y)$, $S$ is mapped to a parallelogram with vertices $(0,0)$, $(3,1)$, $(5,5)$, and $(2,4)$. Using the shoelace formula, the area of this parallelogram is $10$.

The intersection of the image parallelogram and $S$ is the quadrilateral with vertices $(0,0)$, $\left(1, \frac{1}{3}\right)$, $(1,1)$, and $\left(\frac{1}{2}, 1\right)$. To get this quadrilateral, we take away a right triangle with legs $1$ and $\frac{1}{2}$ and a right triangle with legs $1$ and $\frac{1}{3}$ from the unit square. So the quadrilateral has area $1 - \frac{1}{2} \cdot \frac{1}{2} - \frac{1}{2} \cdot \frac{1}{3} = \frac{7}{12}$.

Then the fraction of the image parallelogram that lies within $S$ is $\frac{7}{12} = \frac{7}{120}$, which is the probability that a point stays in $S$ after the mapping.