## Through the HoopProbability puzzles require you to weigh all the possibilities and pick the most likely outcome.
A 1-inch thick steel rod is bent into a perfectly round hoop with an interior diameter of two feet. The hoop is positioned with the opening parallel to the ground. Four inch balls are dropped from above randomly.
Is a ball more likely to hit the hoop or go through the hoop without hitting it? Bonus question: What size hoop would you need in order to have the probability of a ball going through the hoop without hitting it equal the probability of hitting the hoop? ## AnswerThe ball is more likely to hit the hoop.For the ball to either hit the hoop or go through the hoop without hitting it, the center of the ball must be inside of a circle with a diameter equal to the exterior diameter of the hoop plus the diameter of the ball. This is (24 + 2) + 4 = 30 For the ball to go through the hoop without hitting it, the center of the ball must go through a circle with a diameter equal to the interior diameter of the hoop minus the diameter of the ball. This is: (24 - 4) = 20 The probability of the ball going through the hoop instead of hitting it is the area of the smaller circle divided by the area of the larger circle. The area of a circle is Pi*r^2 where r is the radius of the circle, so the probability of the ball going through the hoop instead of hitting it is (Pi * (20/2)^2) / (Pi * (30/2)^2) Pi cancels out of the fraction leaving the probability of going through the hoop without hitting it as: 10^2 / 15^2 = 100/225 = 4/9 = .4444... Bonus question: You need a hoop with a radius r where: 2*(r - 2)^2 = (r + 3)^2 Since both sides must be positive take the square root of each side: sqrt(2)*(r - 2) = r + 3 sqrt(2)*r - 2*sqrt(2) = r + 3 (sqrt(2)-1)*r = 3 + 2*sqrt(2) r = (3 + (2*sqrt(2))) / (sqrt(2)-1) r ~ 14.07 So a hoop with an interior diameter of 2 x 14.07 = 28.14 inches would have an equal chance of the ball hitting the hoop or going through the hoop without hitting it. Hide ## What Next?
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