3 ms·
> A needle of length L is dropped at random onto a sheet of paper ruled with parallel lines a distance L apart. What is the probability that the needle will cro
by maged 11y ago
> A needle of length L is dropped at random onto a sheet of paper ruled with parallel lines a distance L apart. What is the probability that the needle will cross a line?
Thickness of line is needed right? Otherwise P approaches 100% as thickness approaches 0?
- davmre 11y agoThe lines are infinitely thin. Equivalently you can imagine the paper is divided into regions of width L, and the question is whether the needle will cross a region boundary (https://en.wikipedia.org/wiki/Buffon's_needle https://en.wikipedia.org/wiki/Buffon's_needle).
- nonbel 11y agoI don't think that page explains it very well, but have poor math background. I imagined notebook paper with horizontal lines spaced L apart and then the needle dropping at any angle. When the needle is vertical the probability it cross a line is 1, when horizontal it is zero. The length of the needle L is the hypotenuse of a triangle. If we call the angle from horizontal x, the "height" of the needle can be anywhere within h=Lsin(x) for x between 0 and pi/2. The "lines" are like a sample of a point from a uniform distribution U with width L, and h is an interval inside U. The probability a number sampled from a distribution of width L will fall within interval h is h/L. Substituting for h gives p(cross|x) = sin(x). Then assuming the needle is equally likely to drop at any angle, for any one angle theta we get probability density p(theta=x) = 1/(pi/2-0)= 2/pi. The probability the needle drops at angle x AND crosses a line is the product of p(theta=x)p(cross|x)= (2/pi)sin(x). As mentioned, x can range between 0 and pi/2. To get the probability the needle drops at angle x1 OR x2 OR x3, etc and cross we need to sum all these. So take the integral of (2/pi)sin(x) from 0:pi/2. This gives 2/pi.