Solution:
Let $x, y \in [0,1]$. We are to find the probability that $x^{2} - 3 x y + 2 y^{2} > 0$.

We can factor the quadratic:
$$x^{2} - 3 x y + 2 y^{2} = (x - y)(x - 2y)$$
So, $(x - y)(x - 2y) > 0$.

This inequality holds if both factors are positive or both are negative:

Case 1: $x - y > 0$ and $x - 2y > 0$

This gives $x > y$ and $x > 2y$.
But $x > 2y$ implies $x > y$, so the only restriction is $x > 2y$.
But $x \leq 1$, $y \geq 0$, and $x > 2y$.

For $y$ in $[0, 0.5)$, $x$ runs from $2y$ to $1$.

Case 2: $x - y < 0$ and $x - 2y < 0$

This gives $x < y$ and $x < 2y$.
But $x < y$ implies $x < 2y$ (since $y < 2y$ for $y > 0$), so the only restriction is $x < y$.

For $y$ in $[0,1]$, $x$ runs from $0$ to $y$.

So, the region is:
- For $y$ in $[0, 0.5]$, $x$ in $[2y, 1]$ (from Case 1)
- For $y$ in $[0,1]$, $x$ in $[0, y]$ (from Case 2)

The total area (probability) is:

$A = A_1 + A_2$

Where
$$A_1 = \int_{y=0}^{0.5} (1 - 2y) \, dy$$
$$A_2 = \int_{y=0}^{1} y \, dy$$

Compute $A_1$:
$$A_1 = \int_{0}^{0.5} (1 - 2y) \, dy = \left[ y - y^2 \right]_{0}^{0.5} = (0.5 - 0.25) - (0 - 0) = 0.25$$

Compute $A_2$:
$$A_2 = \int_{0}^{1} y \, dy = \left[ \frac{y^2}{2} \right]_{0}^{1} = \frac{1}{2}$$

Therefore, the total probability is:
$$A = 0.25 + 0.5 = 0.75$$

So, the answer is $\boxed{\dfrac{3}{4}}$.