Let $a$ be an integer selected at random from the set $\{0, 1, 2, 3, \ldots, 9\}$. The probability that the equation $ax^2 - ax + 1 = 0$ has real roots is ...

The total number of possible values for $a$ from the set $\{0, 1, 2, 3, \ldots, 9\}$ is $10$.

For the equation $ax^2 - ax + 1 = 0$ to have real roots, we consider two cases:

If $a = 0$, the equation becomes $1 = 0$, which has no roots.

If $a \neq 0$, the equation is a quadratic equation. For it to have real roots, its discriminant must be non-negative.

$D = (-a)^2 - 4(a)(1) \geq 0$

$a^2 - 4a \geq 0$

$a(a - 4) \geq 0$

Since $a \in \{1, 2, 3, \ldots, 9\}$, the condition $a(a - 4) \geq 0$ is satisfied when $a \geq 4$.

The favorable values of $a$ are $4, 5, 6, 7, 8, 9$.

The number of favorable outcomes is $6$.

The required probability is $\dfrac{6}{10} = \dfrac{3}{5}$.

Answer: $\dfrac{3}{5}$