Question

Solve Buffon’s needle problem when L > D. answer: 2L πD(1 − sin θ) + 2θ/π, where cos θ = D/L.

Short answer

The Buffon's needle problem is proved where L > D.

Step-by-step solution

Content Introduction

Given a floor with evenly spaced parallel lines a distance d apart, Buffon's needle problem asks for the probability that a needle of length l would land on a line.

Content Explanation

When L > D, needle will surely cut at least one line for all $\theta$such that $L\cos \left(\theta \right)\ge D$

Therefore from $0to\theta$(such that $\cos {\left(\theta \right)}^1=\frac{D}{L}$needle will surely cut one line at least. Also, will $\theta$be on both sides.

Therefore, out of $\mathrm{\pi }$angle available $2\theta$will give a sure event for angles more than$\theta$.

Explanation Proof

$$\begin{gathered} P\{X<\frac{L}{2}\cos \left(\theta \right)=\iint f_X(x\left)f_0\right(y)dxdy \\ =\frac{4}{\mathrm{\pi D}}{\int }_{{\theta }^1}^{\frac{\mathrm{\pi }}{2}}{\int }_0^{\frac{L}{2}\cos \left(y\right)}dxdy \\ =\frac{4}{\mathrm{\pi D}}{\int }_{\theta }^{\frac{\mathrm{\pi }}{2}}\frac{L}{2}\cos \left(y\right)dy \\ \frac{2L}{\mathrm{\pi D}}(1-\sin \left({\theta }^1\right) \end{gathered}$$

Total probability $\frac{2L}{\mathrm{\pi D}}(1-\sin {\left(\theta \right)}^1+\frac{2\theta }{\mathrm{\pi }}$

where $\cos {\left(\theta \right)}^1=\frac{D}{L}$

Hence proved.