Region R is defined as region in 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?

Probability that \( r > s \) in Region \( R \)

Given: \( R = \{ (x, y) \in \mathbb{R}^2 \mid x^2 + y^2 < 4 \} \) in the first quadrant

Area of region \( R \) in first quadrant: \[ A = \frac{1}{4} \pi (2)^2 = \pi \]

Region where \( r > s \) (i.e., below line \( x = y \)) occupies half of that quarter-circle: \[ A_{\text{favorable}} = \frac{1}{2} \pi \]

Therefore, the required probability is:

\[ \text{Probability} = \frac{\frac{1}{2} \pi}{\pi} = \boxed{\frac{1}{2}} \]