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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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 First Thoughts | Comment 3 of 13 |
Total table area is rectangle of area 100*50
Area where second ball can be placed is reduced to 99*49 (1/2 unit all round) since ball cannot "overhang" the edge.
This must be further reduced by area circle diameter 3 since the second ball cannot be placed (even partially) on top of the first.
Thus total area for placing second ball is (99*49)-(pi*9/4)
"No Go Area" to pot ball is a rectangle length half the diagonal of the table - sqrt(50^2+25^2) and width 3.
So area = 3*sqrt(50^2+25^2)
Probability (as a %) = (3*sqrt(50^2+25^2))/((99*49)-(pi*9/4))
Now the tricky bit. How do I take into account the pocket is twice the width of the ball?

 Posted by Vernon Lewis on 2006-02-11 14:19:30
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