Many mathces of length 2" each were scattered on lined paper?

Question

Parallel lines printed on the paper are separated by distance 2".
Those matches which cross a line were counted, and the number came out 2,000?

How many matches were scattered?

Answer 1

Consider the probability distribution of a match's position between two lines.
Let O be the center point of the match
Let x = distance between between O and the left line
Let θ = angle match makes with horizontal

x = {0, 2}
θ = {-π/2, π/2}

In order for the match not to cross the lines,
x - cosθ > 0
and x + cosθ < 2
or cosθ < x < 2 - cosθ

The probability of any given position is simply
dx/2 * dθ/(π) = dx dθ/(2π)

So it's simply a case of performing the double integrals:
∫ ∫ dx dθ/(2π)
between the limits -π/2 < θ < π/2, cosθ < x < 2 - cosθ

You end up getting
P(matches not crossing) = (2π-4)/(2π)
P(matches crossing) = (4)/(2π) = 2/π

So if 2000 matches touch, 1000π = 3142 matches were scattered.

*EDIT*
I assumed that lined paper means the lines only go in one direction, as opposed to graph paper.

If however you meant graph paper, then you'll find that for any given angle θ, the starting point of the match can occupy an area equal to
4*(1 - cosθ)*(1 - sinθ)
The total area of each square is 4, so
P(match not crossing | θ) = (1 - cosθ)*(1 - sinθ)

P(match not crossing)
= 2/π ∫ (1 - cosθ)*(1 - sinθ) dθ
for 0 < θ < π/2
= (π-3) / π

So P(match crossing) = 3/π
In this case the total no. of matches would be
2000π/3 = 2094

Answer 2

at first i was going to answer 3000. if there's a point exactly in between 2 lines and a circle with radius 2" is drawn, 1 of 3 radii won't cross the lines. then of course i realized that the center of circle can be anywhere between the 2 lines. so maybe the answerer above is right.

answer must have been 3142.

Answer 3

Use Count Buffon's formula
2000/n = 2/pi
n = 1000pi aprox = 3146 mmatches were scattered.