There are two circles in xy-plane whose equations are $x^{2} + y^{2} - 2 y = 0$ and $x^{2} + y^{2} - 2 y - 3 = 0$ . A point (x,y) is choosen at
random inside the larger circle. Then the probability that the point has been taken from the smaller circle is

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1/4

1/16

1/6

1/2

answer is B.

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Detailed Solution

$x^{2} + y^{2} - 2 y = 0 \Rightarrow x^{2} + {(y - 1)}^{2} = 1$

$x^{2} + y^{2} - 2 y - 3 = 0 \Rightarrow x^{2} + {(y - 1)}^{2} = 4$

$P = \frac{Measure of specified part of the region}{Measure of the whole region}$

$P r o b a b i l i t y = \frac{A r e a o f s m a l l e r c i r c l e}{A r e a o f l a r g e r c i r c l e} = \frac{π {(1)}^{2}}{π {(2)}^{2}} = \frac{1}{4}$

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