Hi All,

We're told that the circular center region (the bull’s-eye) has a radius of 3 inches, each successive circle has a radius 3 inches greater than the one before and the largest scoring ring has an area of 153π square inches. We're asked for the probability that an arrow that hits the target hits the bull’s-eye.

To start, we know that the area of the bulls-eye is 9π. To answer the question, we need to figure out the area of the ENTIRE target.

Since each circle's radius is '3 more' than the immediate circle within it, we could potentially 'map out' the area of each ring until we hit 153π square inches. For example:

Bull's-eye = 9π

2nd circle = radius of 6 = 36π - 9π = 27π -- area of 1st ring

3rd circle = radius of 9 = 81π - 36π = 45π -- area of 2nd ring

Etc.

To save some time, we should note that the area of the outer ring is 153π, so the radius of the largest circle would have to be quite a bit bigger than that of these inner circles... If you look at the area of each increasing ring, you'll notice that the area appears to increase by 18π each time....

45π -- radius of 9

63π -- radius of 12

81π -- radius of 15

99π -- radius of 18

117π -- radius of 21

135π -- radius of 24

153π -- radius of 27

Thus, the overall area of the full target is (27^2)π = 729π and the probability of hitting a bull's-eye is 9π/729π = 1/81

Final Answer:

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