Question 275782
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That would depend a great deal on the size of the circles in relation to the size of the rectangle and whether the random falling of objects is constrained to inside the rectangle or not.  Presuming the rectangle as a universal constraint, find the area of each of the circles.  Add the two areas of the circles.  The sum of the areas of the two circles is the numerator of your probability fraction.  The area of the rectangle is your denominator.


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