 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Throwing needles (Posted on 2005-01-19) If I throw 1000 10-centimeter-long needles on to a tiled floor, where each tile is a 10 cm x 30 cm rectangle, approximately how many needles will end up lying across a crack?

You may assume that the widths of both the needles and the cracks are negligible.

 See The Solution Submitted by Sam Rating: 3.6667 (3 votes) Comments: ( Back to comment list | You must be logged in to post comments.) solution | Comment 1 of 9

First regarding crossing the lines spaced at 10-cm:

Each needle might be oriented at any angle, x, with regard to the parallel lines. When the angle is x, the probability that the 10-cm needle will cross one of the lines spaced at 10 cm is sin(x).  Therefore we need to integrate sin(x) over the possible values of x and divide by the length of the interval of integration to get the average.  Each quadrant has the same values, so we can use one quadrant.  This also avoids the problem of having actually to take the absolute value of the sine, as that is what we really need.  So we get

Integ{0 to pi/2} sin(x) dx / (pi/2) = [- cos(x)]{0 to pi/2} / (pi/2) = 2/pi

The probability of lying on one of the 30-cm-spaced lines is 1/3 that for the 10-cm-spaced lines, or 2/(3*pi).

If the events of crossing each type of line were mutually exclusive, we could just add the probabilities.  Instead they are independent.  Since adding the probabilities would add in twice the instances where both occur, we take P(A or B) = P(A) + P(B) - P(A and B), where P(A and B) is the product of these two independent event probabilities:

2/pi + 2/(3*pi) - 4/(3*pi^2) = (6*pi + 2*pi - 4)/(3*pi^2) = (8*pi - 4) / (3*pi^2)

So in 1000 trials you'd get about 714 needles crossing at least one crack. The standard deviation would be about sqrt(714*.386) or 17.

Edited on January 19, 2005, 2:49 pm
 Posted by Charlie on 2005-01-19 14:29:05 Please log in:

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