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(A) Let the radius of the circle be $a$ and the semi-minor axis of the ellipse be $b$. Since the ellipse is inscribed in the circle,the semi-major axi

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$\frac{2\sqrt{2}}{3}$

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(A) Let the radius of the circle be $a$ and the semi-minor axis of the ellipse be $b$. Since the ellipse is inscribed in the circle,the semi-major axis of the ellipse is $a$. The area of the circle is $A_c = \pi a^2$. The area of the ellipse is $A_e = \pi ab$. The probability that a point chosen at random inside the circle lies outside the ellipse is given by $P = \frac{A_c - A_e}{A_c} = 1 - \frac{A_e}{A_c} = 1 - \frac{\pi ab}{\pi a^2} = 1 - \frac{b}{a}$. Given $P = 2/3$,we have $1 - \frac{b}{a} = \frac{2}{3}$,which implies $\frac{b}{a} = \frac{1}{3}$. The eccentricity $e$ of an ellipse is given by $e = \sqrt{1 - \frac{b^2}{a^2}}$. Substituting $\frac{b}{a} = \frac{1}{3}$,we get $e = \sqrt{1 - (\frac{1}{3})^2} = \sqrt{1 - \frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$.

An ellipse is inscribed in a circle and a point within the circle is chosen at random. If the probability that this point lies outside the ellipse is $2/3$,then the eccentricity of the ellipse is:

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