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 11: Two persons $A$ and $B$ agree to meet at a place between 5 and 6 pm. The first one to arrive waits for 20 min and then leave. If the time of their arrival be independent and at random, then the probability that $A$ and $B$ meet is

Accepted Answer

Step 1: Identify the constraint. The ellipse has eccentricity e = 2√2/3 and is inscribed in a circle of radius a. Since inscribed, the semi-major axis A = a.

Step 2: Use eccentricity relation: e² = 1 - b²/a² gives (8/9) = 1 - b²/a², so b²/a² = 1/9, thus b/a = 1/3.

Step 3: Calculate areas. Area of circle = πa². Area of ellipse = πab = πa·(a/3) = πa²/3.

Step 4: The shaded region (outside ellipse but inside circle) has area = πa² - πa²/3 = 2πa²/3.

Step 5: Probability = (Area of shaded region)/(Area of circle) = (2πa²/3)/(πa²) = 2/3.

Note: The solution in the problem uses b directly instead of b/a, but since b = a/3, the result 1 - 1/3 = 2/3 is correct.

∴ Answer: D (p = 2/3)

Question:

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