Question

Two points are selected randomly on a line of length 32 so as to be on opposite sides of the midpoint of the line. In other words, the two points X and Y are independent random variables such that X is uniformly distributed over [0,16) and Y is uniformly distributed over (16,32]. Find the probability that the distance between the two points is greater than 9.

Answer 1

Solution :

Let us consider the squares be $[1, 16] \times [16, 32]$

If x ranges from the 0 to 16 and the y ranges from 16 to 32, we see that the boundary of the region$y-x \geq 9 \text{\ is}\ y - x = 9$  which goes from the $(7, 16 ) \text{ to}\ (16, 25)$.

And so it is easier to find the area of region where $y-x \leq 9$. This is the triangle with points $(7,16),(16,25) \text{ and}\ (16,16)$ as its vertices.

The area if the triangle is =  $\frac{1}{2} \times 9 \times 9$

                                         = $\frac{81}{2}$

Now the entire area is $(16)^2$  = 256

Then, $P(y-x \leq 9) = \frac{81/2}{256} =\frac{81}{512}$

or $P(y-x \geq 9) = 1 - \frac{81}{512}=\frac{431}{512}$

Thus the answer is $\frac{431}{512}$