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 Snooker table (Posted on 2006-02-11)
You are given a 100*50 snooker table (felt area) and two balls of diameter 1. One ball is placed in the center of the table and the other ball is randomly positioned. What is the probability that I will be able to shoot this second ball directly into the top left pocket without touching the central ball? (Assume pocket has radius 1)

 No Solution Yet Submitted by Andre Rating: 4.0000 (2 votes)

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 Solution | Comment 9 of 13 |

Draw two lines between opposite corner pockets. The first goes from the left wall of the upper right pocket (sqrt(2), 0) down to the left wall of the lower left pocket (50, 100-sqrt(2)).

The second line is drawn from the right wall of the upper right pocket (0, sqrt(2)) to the right wall of the lower left pocket (100, 50-sqrt(2)).  These two lines should be parallel and 2 apart.
These lines form the zone where both balls are in some form of alignment from the point of view of the upper right pocket. The half of the zone that falls behind the center ball is the exclusion zone.

Length of each line (d):

d^2 = (x2 - x1)^2 + (y2 - y1)^2
d^2 = (50 - sqrt(2))^2 + (100-sqrt(2) - 0)^2
d^2 = (2500 - 100sqrt(2) +2) + (10000 - 200sqrt(2) +2)
d^2 = 12504 - 300sqrt(2)
d = 109.9
Width of area between lines: w=2
Area between lines: A = d*w =  219.8
Area of exclusion area behind center ball: EA = .5A = 109.9
Additional semicurcular area excluded around center ball.(r = 1 due to distance beteen center of two balls nest to each other)

.5*pi*r^2 = .5*pi

Total exclusion area = 109.9 + .5*pi

Total area = 100 * 50 = 5000

Probability that ball outside exclusion area:

= (5000 - (109.9 +.5pi)) / 5000

= 97.8%

 Posted by Leming on 2006-02-12 13:16:40

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