With a and b positive numbers, a needle of length $l\in(0,min(a,b))$ is dropped randomly on a rectangular grid consisting of an infinite number of parallel lines distance a apart, and perpendicular to these an infinite number of parallel lines distance b apart. Let A and B respectively, be the events that the needle intersects the group of lines at distance a and b apart.
Show that $P(A)=\frac{2l}{a\pi}$ and $P(B)=\frac{2l}{b\pi}$ and determine $P(A\cap B)$ and verify that A and ? Are strictly negatively correlated, that is; $P(A\cap B)<P(A)P(B)$.
I assumed wlog the horizontal lines are a apart and vertical ones are b apart. Defined $\theta$ to be the r.v. that represents the angle between the needle and horizontal lines. So, $\theta$ had distribution $Uniform[0,\frac{\pi}{2})$ by symmetry. Then defined X as the distance between the midpoint of the needle and horizontal lines and Y to be the distance between the midpoint of the needle and the vertical lines. So, again by symmetry, X has distribution $Uniform[0,\frac{a}{2})$ and Y has distribution $Uniform[0,\frac{b}{2})$. Then $P(A)=P(\frac{X}{sin(\theta)}<\frac{l}{2})$ and integrating I obtained the desired result. Then similarly I calculated $P(B)$. My problem is the second part, that is finding $P(A\cap B)$. I think $P(A\cap B)=P(\frac{X}{cos(\theta)}<\frac{l}{2}, \frac{Y}{sin(\theta)}<\frac{l}{2})$ and cannot calculate this probability.
Best Answer
You should have
\begin{eqnarray*} P(A\cap B) &=& P\left(\dfrac{X}{\sin\theta} \lt \dfrac{l}{2} \bigcap \dfrac{Y}{\cos\theta} \lt \dfrac{l}{2}\right) \\ &=& \int_{\theta=0}^{\pi/2}\int_{x=0}^{\frac{l}{2}\sin\theta}\int_{y=0}^{\frac{l}{2}\cos\theta} \dfrac{2}{\pi} \dfrac{2}{a} \dfrac{2}{b} \;dy\;dx\;d\theta \\ \end{eqnarray*}
Completing that integral should give you the required result.