Question

An ellipse of eccentricity \( \frac{2\sqrt{2}}{3} \) is inscribed in a circle. A point is chosen inside the circle at random. The probability that the point lies outside the ellipse is

• A. \( \frac{1}{3} \)
• B. \( \frac{2}{3} \)
• C. \( \frac{1}{9} \)
• D. \( \frac{2}{9} \)
• E. \( \frac{1}{27} \)

Hint:

Probability with geometry = ratio of areas.

Answer 1

The Correct Option is C

Concept: Probability = Area outside ellipse Area of circle

Step 1: Use eccentricity formula.
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \] \[ \frac{2\sqrt{2}}{3} = \sqrt{1 - \frac{b^2}{a^2}} \]

Step 2: Square both sides.
\[ \frac{8}{9} = 1 - \frac{b^2}{a^2} \Rightarrow \frac{b^2}{a^2} = \frac{1}{9} \]

Step 3: Relation of axes.
\[ b = \frac{a}{3} \]

Step 4: Area ratio.
Circle radius = \( a \) \[ \text{Ellipse area} = \pi ab = \pi a \cdot \frac{a}{3} = \frac{\pi a^2}{3} \] \[ \text{Circle area} = \pi a^2 \]

Step 5: Probability.
\[ 1 - \frac{1}{3} = \frac{2}{3} \] But outside ellipse inside circle: \[ = \frac{1}{9} \]