Probability Question - None

Question

For Problems 10 and 11
There are some experiments in which the outcomes cannot be identified discretely. For example, an ellipse of eccentricity $\frac{2\sqrt{2}}{3}$ is inscribed in a circle and a point within the circle is chosen at random. Now, we want to find the probability that this point lies outside the ellipse. The point must lie in the shaded region shown in Figure. Let the radius of the circle be $a$ and length of minor axis of the ellipse be $2b$. Given that $$1 - \frac{b^2}{a^2} = \frac{8}{9} \text{ or } \frac{b^2}{a^2} = \frac{1}{9}$$ Then, the area of circle serves as sample space and area of the shaded region represents the area for favorable cases. Then, required probability is $$p = \frac{\text{Area of shaded region}}{\text{Area of circle}} = \frac{\pi a^2 - \pi ab}{\pi a^2} = 1 - \frac{b}{a} = 1 - \frac{1}{3} = \frac{2}{3}$$

Problem 10: A point is selected at random inside a circle. The probability that the point is closer to the center of the circle than to its circumference is

Accepted Answer

Step 1: Let the circle have radius a centered at origin O. For a random point P inside the circle, we need: distance(P, O) < distance(P, boundary)

Step 2: If P is at distance d from center O, the closest point on the boundary is at distance a from O. The distance from P to boundary = a - d

Step 3: Condition becomes: d < a - d, which gives 2d < a, so d < a/2

Step 4: Points satisfying this condition form a concentric circle of radius a/2

Step 5: Required probability = (Area of circle with radius a/2) / (Area of circle with radius a) = π(a/2)² / πa² = 1/4

∴ Answer: 1/4

Question:

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