Probability
Probability
star_batch_jee_advanced_2025
Grade 12
Question:
A point is selected at random inside an equilateral triangle whose side length is 3. The probability its distance to any corner is greater than 1 is
\frac{2\pi}{9\sqrt{3}}
1 - \frac{2\pi}{9\sqrt{3}}
\frac{\sqrt{3}\pi}{9}
1 - \frac{\sqrt{3}\pi}{9}
Step-by-Step Solution
Key Concept: Subtract the unfavorable region (circular segments) from the total triangle area to find probability.
The area of an equilateral triangle with side 2 is $\frac{\sqrt{3}}{4}(3)^2 = \frac{9\sqrt{3}}{4}$. Points must lie in the shaded region. The area of each circular segment is $\frac{\pi}{6}(1)^2$. The desired probability is $1 - \frac{3\pi}{4\sqrt{3}} - \frac{2\pi}{9\sqrt{3}}$, which accounts for the three circular segments of radius 1 centered at each vertex.
Correct Answer: 2