Somewhat challenging question (1 Viewer)

[icycloud]

This took me a while to work out. I thought the question was pretty interesting, so decided to share it with everyone:

A 2cm needle is dropped at random onto a large piece of paper with lines ruled parallel at 2cm apart. Find the probability that the needle misses all the lines. Treat the needle as a line segment.

 

[SeDaTeD]

Is it 1 - 2/pi?

I'll put working later, in case someone else wants to have a dig.

 

[icycloud]

Yup, that's what I got. I'd be interested to see your working Use the spoiler tag perhaps?

 

[Riviet]

I'm not sure of how to even start this question. XD

Any hints guys?

 

[mitsui]

wow! is it...in the hsc course?! (plz say no... )

 

[Sober]

Spoiler

Imagine the lines are horizontal, we want to find the vertical distance (v) from the bottom of the pin to the top (how far it spans between the lines). The chances of it landing on a line is (2 - v)/2.

The value of v will be the 2sin(theta), but since we want an average we take the integral from zero to pi then divide by pi and we get 4/pi:

(2-4/pi)/2 = 1-2/pi

 

[Riviet]

wow! is it...in the hsc course?! (plz say no... )

Nah it's not, otherwise Icycloud wouldn't have posted it in this forum.

Quit interesting too, since it combines probability, trig and integration into one.

 

[icycloud]

Nice one Sober. Here was my method (somewhat more drawn out than Sober's):

Spoiler

Consider only one line. Now, the center of the needle can land within 1cm of the line, above or below. Let the vertical distance from the center to the line be 'x', and the angle between the needle and 'x' be theta [@]. [See diagram]

Now, from the diagram we have

@ = arccos(x)

The needle intersects the line when 0 < @ < arccos(x)
Thus, the probability of the needle intersecting is arccos(x)/pi {max theta = pi}
However, the needle centre can be above or below, so we multiply by 2:

P(needle intersects) = 2arccos(x)/pi
P(needle misses) = 1-2arccos(x)/pi

But 0<x<1, so we integrate over 0-->1,

Int[0-->1] (1-2arccos(x)/pi) dx
= 1 - 2/pi #

{arccos integrated by parts}

 

[SeDaTeD]

Spoiler

Well I considered the probability of the centre of the needle landing in a strip between x and x+dx above a line. Then I found the probability of the needle crossing the line in that case, taking it to be centred at x away from a line. The summation of all these bands give an approximation of the total probability. Then i took the limit as dx->0 whihc led to an integral : 1 - INT 0->pi/2 {(2/pi)*theta*sin(theta) dtheta} - where theta is the angle between the needle and the perpendicular to the lines. (I know I should have chose it to be the one between the needle and a line parallel to the lines, but same result). Evaluating the integral gives 1 - 2/pi.

 

[who_loves_maths]

the famous Buffon's Needle problem is one of the "extension" questions in the Probability chapter of the 3u Cambridge (either the Yr11 or Yr12 book, i forgot which one exactly) book.

interestingly, there have been real life experiments of the Buffon Needle scenario used to provide a "practical" estimation of the value of pi.

continuous probability is always a good area to see where/how mathematics can be directly applied to the physical world

 

[Templar]

interestingly, there have been real life experiments of the Buffon Needle scenario used to provide a "practical" estimation of the value of pi.

If you want to toss a needle a couple million times and still don't get near the third convergent, sure, it's practical.

 

[who_loves_maths]

^ exactly, and that's why the word 'practical' was put in inverted commas in my quote Templar

Oh and one more thing - only dumb experimenters would ever even think of doing it manually. the smart ones usually use a little thing we call 'computer'