Question 187968
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Presuming the circles are also touching the sides of the rectangle and that the circles are all the same radius, then the length of the rectangle is 3 times the diameter or 6 times the radius of one of the circles and the width of the rectangle is 2 times the diameter or 4 times the radius of one of the circles.  So the area of the rectangle, in terms of the radius of one of the circles is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6r\,\cdot\,4r = 24r^2]


The area of one of the circles is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \pi r^2]


So the area of all of the circles is 6 times that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6\pi r^2]


Again, I have to make an assumption because you didn't say.  Assuming that the darts are constrained to the rectangle, meaning that the probability of any given dart hitting somewhere in the rectangle is 1, then the probability of a dart landing in a circle is the area of all of the circles divided by the area of the rectangle:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{6\pi r^2}{24r^2} = \frac{\pi}{4}]


Giving you the exact answer.  The approximate numercial value would depend on the value used for &#960;.  However it is safe to state:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.78 < p < 0.79]




John
*[tex \LARGE e^{i\pi} + 1 = 0]
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