To find the probability that $BD < \sqrt{2}$, we can analyze the geometric properties of triangle $ABC$.
Since $\angle C = 90^\circ$, triangle $ABC$ is a right triangle. Given that $\angle B = 60^\circ$, we can determine the lengths of the sides using trigonometric ratios.
Let's denote $AC$ as $c$ and $BC$ as $a$. We have $AB = 2$.
Using the properties of a $30^\circ$-$60^\circ$-$90^\circ$ triangle, we know that the side opposite the $30^\circ$ angle is half the length of the hypotenuse. Thus, $AC = c = 2 \cdot AB = 4$.
Now, let's consider point $D$ on side $AC$ such that $BD < \sqrt{2}$. We need to find the range of positions for $D$ that satisfy this condition.
Since $\triangle ABC$ is a right triangle, we can see that the maximum length of $BD$ occurs when $D$ is at point $C$. In this case, $BD = BC = a$.
We want to find the range of $a$ such that $BD < \sqrt{2}$.
From the Pythagorean theorem, we have $BC^2 + AB^2 = AC^2$. Substituting the given values, we get $a^2 + 2^2 = 4^2$, which simplifies to $a^2 + 4 = 16$. Solving for $a$, we find $a = 4\sqrt{3}$.
Therefore, the range of values for $a$ such that $BD < \sqrt{2}$ is $0 < a < 4\sqrt{3}$.
To calculate the probability, we divide the length of the desired range by the total possible length of $a$, which is $0 < a < 4$ (since $0 < a < 4$ covers the entire base $AC$ of the triangle).
The probability is thus $\frac{4\sqrt{3}}{4} = \sqrt{3}$.
Therefore, the probability that $BD < \sqrt{2}$ is $\sqrt{3}$.
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