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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.

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  • $\begingroup$ You are right about your density function and that you must find a certain domain over which to integrate. Calculate the probability that the distance is at most 6. If $y<9$ then we know that it is impossible to find a value of $x$ satisfying the distance of at most 6. Now, for a fixed value of $y$ between $9$ and $15$, what are the values of $x$ which are acceptable? We know that $x>15$, what is the upper bound on $x$ for the value of $x$ to be acceptable? This should depend on $y$. $\endgroup$
    – tjeremie
    Dec 10, 2019 at 3:38

1 Answer 1

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Instead of integration, it may be simpler to take a more "graphical" approach!

For example,

  1. 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 $$\{\left(x,y\right)\in\mathbb{R}^2 : 0 \le x \le 15 \text{ and } 15 \le y \le 30\}$$

  2. 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.

  3. 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 $$A_\text{square} =15 \cdot 15 = 225$$

  4. 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 $$A_\text{triangle} = 6 \cdot 6 \div 2 = 18$$

  5. 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!

  • Edit: I suppose if you must use "integration", you could always express the probability like $$P\left(y-x>6\right) = P\left(x<9\right) + P\left(x>9\right)P\left(y-x>6 \;\big|\; x > 9\right) = \frac{9}{15} + \frac{6}{15} \left(\frac{1}{15-9}\int_9^{15}\frac{30-\left(x+6\right)}{15}dx \right) = \boxed{\frac{23}{25}}$$
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