- #1

LCSphysicist

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- TL;DR Summary
- How you would approach this problem? We need to find the accesibles position of a dropping needle

My approach to this problem is a little laborious, it involves three coordinates, probably it is right, but tiring and extensive beyond what the question wanted.

Be the origin in the rectangle middle.

It would be like: imagine a rectangle with opposite sides L and R with length l, so to find the probability the needle cross a line, i would at first find the probability the needle center fall in a position between x,x + dx; y, y + dy, so find the allowed angle to it rotate without cross a line, find the probability it falls in this range (range/2pi), ant integrate wrt dx,dy, finding Pf. But, as what we want is exactly opposite of it, we would just do P = 1 - Pf

In another words:

$$P = 1- \int\int P((x,y))P(\theta)dxdy$$

where we can express $$\theta = \theta(y)$$ (what matters to the angle is y itself, with it we can know the distance between the line and the needle center as l-y

In the end the x value will cancel $P(x,y) = \frac{dxdy}{lX}$, when we integrate dx, it will just cancel with X) and we will finish with just y.

But i believe maybe there is an easier way to do it... What would you do?