Two points are selected randomly on a line of length 30 so as to be on opposite sides of the midpoint line. In other words, the two points X and Y are independent random variables such that X is uniformly distributed over [0,15) and Y is uniformly distributed over (15,30].
Find the probability that the distance between the two points is greater than 6.
P( |X-Y| > 6)= 1 - P( |X-Y| < 6)=?
I am not sure where to start, I that f(x , y) = 1/225, and that I am supposed to set up a double integral to integrate this function. I am not sure what to make my bounds however.
Instead of integration, it may be simpler to take a more "graphical" approach!
For example,
Let $x$ and $y$ be your two random variables of interest. We can see that the set of all possible points $\left(x,y\right)$ can be represented as a square in the Cartesian plane $$\{\left(x,y\right)\in\mathbb{R}^2 : 0 \le x \le 15 \text{ and } 15 \le y \le 30\}$$
We notice the line $y = x+6$ intersects this square, and $y-x>6$ for all points above this line, while $y-x<6$ for all points below.
What we are looking for is the probability $P\left(y-x>6\right)$, which can be simply represented as the ratio of the area of the square above this line to that of the whole square. Since the square has a side length of $15$, we know its area to be $$A_\text{square} =15 \cdot 15 = 225$$
We can also see graphically that the area below this line in the square is represented as a right triangle with base and altitude both equal to $6$. Thus, its area must be $$A_\text{triangle} = 6 \cdot 6 \div 2 = 18$$
Now, we can simply solve for the desired probability as $$P\left(y-x>6\right) = 1 - P\left(y-x \lt 6\right) = 1 - \frac{A_\text{triangle}}{A_\text{square}} = \boxed{\frac{23}{25}}$$
Hope that helps!