Given 2 real numbers x and y such that 0 ≤ x,y ≤ 2,find the probability that x2 + y2 ≤ 1.
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1 Answers
66All possibilities for x,y is the square having its vertices at (0,0), (2,0), (2,2), (0,2). And the favorable conditions are the set of all points (x,y) lying in the first quadrant within the circle x2 + y2 = 1. Hence, the required probability is the ratio of the areas of the quadrant of the circle to that of the square:
i.e. \dfrac{\pi/4}{4}=\dfrac{\pi}{16}