In a right angled $\triangle ABC$, $DE$ and $DF$ are perpendicular to $AB$ and $BC$ respectively. What is the probability of $DE\cdot DF>3$?

Let $DC = x , DF = o , DE = a $ and $ DA=y $ .

We have :- $$ o = x \sin 15 \tag{1}$$ $$a= y \ cos 15 \tag{2} $$ Multiplying $(1),(2)$ , we get :- $$ o\cdot a = xy \sin 15 \cos 15 = \frac{xy}{2} sin 30 = \frac{xy}{4} > 3 $$ ( $\because 2 \sin \theta \cos \theta = \sin 2\theta $)

Hence , we need $xy = x(7-x) > 12$ or $x^2-7x+12<0$

As this is of the form $ax^2+bx+c$ , and since $a>0$ , the expression is negative between the roots . Therefore, we must have $x \in (3,4) $

$\therefore $ The probability $\frac{4-3}{7} = \frac{1}{7}$

Here is how one can solve the problem with a visualization of the issue, i.e., with a display of the line segment where $D$ must be situated in order that the area condition is fulfilled. Let $(x,y)$ be the coordinates of $D$ with respect to the natural axes of the figure. As the constraint is $xy>3$, the limit is provided by the hyperbola with equation $xy=3$. The coordinates (x_1,y_1) and (x_2,y_2) of intersection points $D_1$ and $D_2$ are thus solutions of the following system :

$$\begin{cases}\frac{x}{s}&+&\frac{y}{s}&=&7\\ &xy&&=&3\end{cases} \ \ \ \text{where} \ s:=\sin(15°) \ \text{and} \ c:=\cos(15°)$$

We have thus transformed our query into a classical problem : we will get a quadratic equation, out of which we will obtain

$$D_1=(\frac{1}{s},\frac{3}{4c}) \ \text{and} \ D_2=(\frac{3}{4s},\frac{1}{c})$$

giving length $D_1D_2=1$. The final answer is $D_1D_2/AC=1/7$ : we find back the same result as @Rahuboy.