Let the length of the needle be $l$, the distance between the lines be $d$.
Suppose $l>d$.
Let $X$: the distance from the midpoint of the needle to the nearest line,
and $\Theta$: the angle of the needle with respect to the line. ($0 \le \Theta \le \frac{\pi}{2}$)
Let $C$ be the event when the needle crosses a line.
If we let, $\sin {\alpha} = \frac{d}{l}$,
$$
P[C] = P[C|\Theta\ge \alpha]P[\Theta \ge \alpha]+P[C|\Theta \lt \alpha]P[\Theta \lt \alpha]
$$
$$
P[\Theta \ge \alpha] = \frac{2}{\pi}\left(\frac{\pi}{2}-\alpha\right), P[\Theta \lt \alpha]=\frac{2}{\pi}\left(\alpha\right)
$$
$$
P[C|\Theta\ge \alpha]=1
$$
$$ P[C|\Theta \lt \alpha]=P[X<\frac{l}{2}\sin{\Theta}]=\int_{0}^{\alpha} \int_{0}^{\frac{l}{2}\sin{\theta}}\frac{4}{d\pi}dxd\theta
$$
This gives a wrong answer. I cannot find out what I did wrong.
Best Answer
Your final integral $$ \int_{0}^{\alpha} \int_{0}^{\frac{l}{2}\sin{\theta}}\frac{4}{d\pi}dxd\theta\tag1 $$ is computing $P[\{X<\frac l2\sin\Theta\}\cap\{ \Theta<\alpha\}]=P[C \cap \{\Theta< \alpha\}]$, not the conditional probability $P[C\mid \Theta< \alpha]$. This means that (1) is already $P[C\mid \Theta<\alpha]P[\Theta<\alpha]$, and you can stop there; you don't want to multiply (1) by $P[\Theta<\alpha]$.