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vhbelvadi

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**Q. A box has a square bottom 1m on a side, vertical side walls 0.33m tall and no top. The box is set so its bottom is accurately horizontal, and 7 tennis balls are thrown into it. One more such ball is dropped vertically into the box. What is the probability that the dropped ball hits the bottom of the box without encountering one of the balls already in the box?**

source: http://www.feynmanlectures.info/FLP_Original_Course_Notes/pages/ON1-215.html

**2. A rather vague attempt at a solution:**Well, right now, I haven't gone too far from the basic probability equation (P(G) being the probability of hitting the ground, r being the radius of a tennis ball in metre):

[itex]P(G) = \frac{1m^2-\{7*\pi r^2\}}{1m^2} = 1-7*\pi r^2[/itex]

(What with the bottom being perfectly horizontal and all (normal distribution?), I thought of working along the lines of the coin-tile problem here: http://mathworld.wolfram.com/CleanTileProblem.html )

But I doubt this is correct: 1. wouldn't we need to know the diameter of a tennis ball? Or, if we can work it out in terms of

*r*, 2. should the vertical wall length not have something to do with it? Or am I going in a completely wrong direction?