Buffon’s needle

The plane is ruled by parallel lines $2$ inches apart and a $1$-inch long needle is dropped at random on theplane. What is the probability that it hits parallel lines?
Solution.
The first issue is to find some appropriate probability space $(\mathrm{\Omega },ℱ,P)$. For this,

• $h=$ distance from the center of the needle to the nearest line

• $\theta =$ the angle that the needle makes with the horizontal ranging from $0$ to $\frac{\pi }{2}$.

These fully determine the position of the needle. Let us next take the

1. The probability space is $\mathrm{\Omega }=[0,1]\times [0,\frac{\pi }{2})$

2. The probability of an event $B$ is denoted by $P[B]$ is equal to$\frac{areaofB}{\frac{\pi }{2}}$

Now we denote by $A$ the event that the needle hits a horizontal line. It is easily seen that this happenswhen $\sin \theta \ge \frac{h}{1/2}$. Consequently $A=\{(\theta ,h)\in \mathrm{\Omega }:h\le \frac{\sin \theta }{2}\}$and then we get $P[A]=\frac{2}{\pi }{\int }_0^{\frac{\pi }{2}}\frac{1}{2}\sin \theta d\theta =\frac{1}{\pi }\mathrm{\square }$

In general case, when the length of needle is $l$ and the distance of parallel lines is $d$ provided that , the probability we want is $\frac{2l}{\pi d}$. This is obvious just taking the $l/d$-point from one edge instead of the center of the needle.