Is my solution to Buffon's needle OK?

[SamRoss]

TL;DR

My solution is so much simpler than the solution provided that I'm doubting myself.

Buffon's needle was presented as a problem in David Griffiths' "Introduction to Quantum Mechanics". In the book, a needle is of length l is dropped randomly on a sheet of ruled paper with the lines of the paper also a distance l apart. It is required to find the probability of the needle crossing one of the lines. I reasoned as follows...

P (needle crosses a line) = P (angle between the needle and a perpendicular to the lines is Θ) * P (the needle crosses a line, given the angle Θ)

Taking symmetry into account, I only considered Θ taking on values from 0 to π/2. Then the probability that Θ ends up being between Θ and Θ+dΘ is $\frac {dΘ} {π/2} = \frac 2 π dΘ$.
Next, also taking symmetry into account, I only considered the space between one pair of the ruled lines. The projection that the needle makes with the perpendicular is $l\cosΘ$ and since the lines are also l apart, the probability, given Θ, that the needle crosses a line is $\frac {l\cosΘ} {l}=\cosΘ$. So...
P (needle crosses a line) = $\frac 2 π\int_0^{π/2} \cosΘ \, dΘ=\frac 2 π$.

This is the answer shown on the online answer key, but the explanation is so much longer I thought maybe there was a mistake in my reasoning and I only arrived at the right answer by coincidence. This happens to my students sometimes (I teach middle school math). They'll make mistake after mistake while solving an equation and magically end up with the correct solution. I want to be sure the same thing is not happening to me.

 

[scottdave]

I had to read it twice to visualize what you're talking about. But it looks valid to me.

 

[PeroK]

Summary:: My solution is so much simpler than the solution provided that I'm doubting myself.

Buffon's needle was presented as a problem in David Griffiths' "Introduction to Quantum Mechanics". In the book, a needle is of length l is dropped randomly on a sheet of ruled paper with the lines of the paper also a distance l apart. It is required to find the probability of the needle crossing one of the lines. I reasoned as follows...

P (needle crosses a line) = P (angle between the needle and a perpendicular to the lines is Θ) * P (the needle crosses a line, given the angle Θ)

Taking symmetry into account, I only considered Θ taking on values from 0 to π/2. Then the probability that Θ ends up being between Θ and Θ+dΘ is $\frac {dΘ} {π/2} = \frac 2 π dΘ$.
Next, also taking symmetry into account, I only considered the space between one pair of the ruled lines. The projection that the needle makes with the perpendicular is $l\cosΘ$ and since the lines are also l apart, the probability, given Θ, that the needle crosses a line is $\frac {l\cosΘ} {l}=\cosΘ$. So...
P (needle crosses a line) = $\frac 2 π\int_0^{π/2} \cosΘ \, dΘ=\frac 2 π$.

This is the answer shown on the online answer key, but the explanation is so much longer I thought maybe there was a mistake in my reasoning and I only arrived at the right answer by coincidence. This happens to my students sometimes (I teach middle school math). They'll make mistake after mistake while solving an equation and magically end up with the correct solution. I want to be sure the same thing is not happening to me.

You may need just a bit more justification, but I think it's a nice shortcut method.