Let $$A$$ be the event that the absolute difference between two randomly chosen real numbers in the sample space $$[0, 60]$$ is less than or equal to $$a$$. If $$P(A) = \dfrac{11}{36}$$, then $$a$$ is equal to ______.

Two real numbers $$x$$ and $$y$$ are chosen randomly from $$[0, 60]$$, so the sample space is a square of side 60 whose area is $$60^2 = 3600$$, and the event $$A$$ defined by $$|x - y| \leq a$$ corresponds to the region between the lines $$y = x + a$$ and $$y = x - a$$ within this square.

Since the area outside this region consists of two right triangles each with legs of length $$(60 - a)$$, the probability of $$A$$ is the complement of the ratio of their total area to 3600, namely:

$$ P(A) = 1 - \frac{(60 - a)^2}{60^2} = \frac{3600 - (60 - a)^2}{3600} = \frac{120a - a^2}{3600}. $$

Imposing the condition $$P(A) = \tfrac{11}{36}$$ leads to:

$$ \frac{120a - a^2}{3600} = \frac{11}{36}, \quad 120a - a^2 = 1100, \quad a^2 - 120a + 1100 = 0. $$

By the quadratic formula one finds:

$$ a = \frac{120 \pm \sqrt{14400 - 4400}}{2} = \frac{120 \pm \sqrt{10000}}{2} = \frac{120 \pm 100}{2}, $$

so that $$a = 110$$ or $$a = 10$$, and since $$a$$ must lie within $$[0,60]$$, it follows that $$a = 10$$.