1
$\begingroup$

A needle 2.5cm long is dropped on a piece of paper that has a very fine parallel lines 2.25cm apart drawn on it.

What is the probability that the needle lies between the two lines?

I can see how this would work if the length of the needle was less than the length of the lines but since it is the other way around I get a little lost. If anyone could explain this thoroughly I would appreciate it highly.

Thank you

Edit:

Well I can see that the 2 chance variables: the midpoint of the needle's least distance from a like so say that variable is x and x is between 0 and 1.125, and the other chance variable which is the rotation of the needle. Call this rotation Q and Q is between 0 and pi/2.

This is where I start to get muddy and unsure. So the needle should not cross the line if x>(1.25*Sin(Q)). So This outside integral of the double integral integrates in terms of Q from 0 to (pi/2) the inside integral I'm not sure from what to what.

The function being integrated is (16pi)/9 as it is just the probability density functions multiplied. Can anyone help me at this point?

$\endgroup$
1
$\begingroup$

This is just a hint, these sorts of problems are more fun if you work it out for yourself. In Buffon's needle the lines are parallel, but the needle is assumed to be rotated uniformly.

So the angle between the needle and the lines is assumed to be uniformly distributed between $0$ and $90^\circ$. So if the needle is nearly parallel to the lines you have a high probability that it lands between them.

If you've worked this out already and you're stuck on some detail then post a comment and modify your question with how far you've got and I (or someone else) will help. But this observation might be enough if that was what was confusing you.

$\endgroup$
2
  • $\begingroup$ Well I can see that the 2 chance variables. the midpoint of the needle's least distance from a like so say that variable is x and x is between 0 and 1.125. The other chance variable which is the rotation of the needle. Call this Q and Q is between 0 and pi/2. $\endgroup$ – Nicholas May 7 '13 at 14:14
  • $\begingroup$ This is where I start to get muddy. So the needle should not cross the line if x>(1.25*Sin(Q)). This outside integral of the double integral integrates in terms of Q from 0 to (pi/2) the inside integral I'm not sure from what to what. The function being integrated is 16pi/9. $\endgroup$ – Nicholas May 7 '13 at 14:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.