Darts on a ruler probability If two points are selected at random on an interval from 0 to 1.5 inches, what is
the probability that the distance between them is less than or equal to 1/4"?
 A: Draw the square with corners $(0,0)$, $(1.5.0)$, $(1.5,1.5)$, and $(0,1.5)$.
Imagine the points are chosen one at a time. Let random variable $X$ be the first chosen point, and $Y$ the second chosen point. We are invited to assume that $X$ and $Y$ are uniformly distributed in the interval $[0,1.5]$ and independent. (Uniform distribution is highly implausible with real darts.) 
Then $(X,Y)$ is uniformly distributed in the square just drawn.
Consider the two lines $y=x+\frac{1}{4}$ and $y=x-\frac{1}{4}$. 
The two points are within $\frac{1}{4}$ inch from each other if the random variable $(X,Y)$ falls in the part of our square between the two lines.
Call that part of the square $A$. Then our probability is the area of $A$ divided by the area of the whole square. 
Remark: It is easier to find first the area of the part of the square which is not in $A$. This consists of two isosceles right triangles with legs $\frac{5}{4}$, so their combined area is $\frac{25}{16}$. The area of the whole square is $\frac{9}{4}$, so the area of $A$ is $\frac{11}{16}$. 
Thus our probability is $\dfrac{\frac{11}{16}}{\frac{9}{4}}$.
A: I'm assuming the points are selected independently and using the uniform distribution on $[0, 1.5]$.  If you are unfamiliar with these terms you are probably assuming them implicitly and I wouldn't worry about it.  What I would do is draw a square 1.5 units on each side and draw the subset of the square of points $(x,y)$ with $|x-y| \leq 1/4$.  This region will be a hexagon.  Then divide the area of the hexagon by the area of the whole square.
Remark:
It looks like I am competing with another respondent!  I just read his suggestion, and I echo it: find the area of what is NOT in the hexagon, and subtract it from the square, instead of finding the area of the hexagon directly.  Since I stole from his answer, feel free to accept his instead of mine.
A: The square $S=\{(x,y) | x,y \in [0, \frac{3}{2}] \}$ has area $(\frac{3}{2})^2$. The area $\Delta= \{(x,y) \in S \, |\,  |x-y| > \frac{1}{4} \}$ can be easily rearranged to be a square with area $(\frac{3}{2}-\frac{1}{4})^2$. Hence the chance of landing in $S \setminus  \Delta$ is $\frac{(\frac{3}{2})^2-(\frac{3}{2}-\frac{1}{4})^2}{(\frac{3}{2})^2} = 1 - (1-\frac{1}{6})^2 = \frac{11}{36}$.
