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 \(r > s\)?
Step-by-step Solution:
Given Data:
The region \( R \) is the first quadrant satisfying:
\[
x^2 + y^2 < 4
\]
This represents a quarter-circle of radius 2 centered at the origin in the first quadrant.
A random point \( P = (r, s) \) is chosen from \( R \).
We need to find the probability that \( r > s \).
Understanding the Condition \( r > s \):
The line \( x = y \) divides the first quadrant into two equal parts inside the quarter-circle.
Any point above the line \( x = y \) satisfies \( r < s \), and any point below satisfies \( r > s \).
Probability Calculation:
Since the quarter-circle is symmetric about the line \( x = y \), the probability of a randomly chosen point falling in the region where \( r > s \) is exactly half of the total area.
Thus, the probability that \( r > s \) is:
\[
\frac{1}{2}
\]
Final Answer:
Option C: \( \frac{1}{2} \)