Q.6.4 - A First Course in Probability

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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 $0 to \theta$ (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 $\theta$.

Step 3: 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.