Question

Region \( R \) is defined as the region in the first quadrant satisfying the condition \( x^2 + y^2<4 \). Given that a point \( P = (r, s) \) lies in \( R \), what is the probability that \( r>s \)?

• A. \(1\)
• B. \(0\)
• C. \(\frac{1}{2}\)
• D. \(\frac{1}{3}\)

Hint:

In problems involving symmetry and uniform probability distribution over geometric regions, use geometric arguments rather than integration when possible.

Answer 1

The Correct Option is C

Step 1: Understanding the region \( R \). 
The inequality \( x^2 + y^2<4 \) represents a circle of radius 2 centered at the origin, but restricted to the first quadrant (i.e., \( x>0, y>0 \)). 
Step 2: Find total area of region \( R \). 
The total area of a circle is \( \pi r^2 = \pi (2)^2 = 4\pi \), so the first quadrant portion is: \[ {Area of } R = \frac{1}{4} \cdot 4\pi = \pi \] 
Step 3: Consider the line \( x = y \) within region \( R \). 
The line \( x = y \) divides the circular region into two symmetric parts within the first quadrant: one with \( x>y \), and the other with \( y>x \). So the region where \( r>s \) (i.e., \( x>y \)) occupies exactly half of region \( R \). 
Step 4: Compute the required probability. \[ {Probability} = \frac{{Area where } x>y { in } R}{{Total area of } R} = \frac{1}{2} \]