# Geometric probability question

So there are 2 parallel lines 20 feet apart. A piece of pipe 20 feet long falls between the lines and one end is exactly 10 feet from one line. What is the probability that the pipe lies entirely between the lines?

I know how to do this if the pipes placement varies but not if it is fixed. Can anyone help me figure out the function and how I must integrate this?

Thank you

• The answer will depend, as in Bertrand's paradox, in what is meant by falling at random. One assumption that could be made is that the location of the other end is uniformly distributed on the circle of radius 20 ft and center equidistant from the two lines, or as @Umakant suggests, center 10 ft from one line and 30 ft from the other, in which case, the probability of falling between the lines is $0$. – Dilip Sarwate May 8 '13 at 2:42

You just need to compute the range of angles over which the pipe is inside the two libes and then divide by $2 \pi$. It should be clear that thus the angular range is $[\arccos{(1/2)},\pi-\arccos{(1/2)}]$, or a range of $\pi - 2 \pi/3 = \pi/3$. Thus the probability is, by symmetry,
$$\frac{2 \pi/3}{2 \pi} = \frac13$$
Since one end of the pipe is 10 feet from one line. (I am assuming it is in between the two lines not outside. You can think of this also.) If the pipe is parallel to the lines then it's fine. Now start rotating the pipe, till it hits one end. Figure out the total angle during which it stays within the lines and divide by $2\pi$.