\( P(X=0 \text{ and } Y=0) = \frac{1}{4}, \) \( P(X=1 \text{ and } Y=0) = \frac{1}{8}, \) \( P(X=0 \text{ and } Y=1) = \frac{1}{2}, \) \( P(X=1 \text{ and } Y=1) = \frac{1}{8}. \)
Consider two statements: Statement 1: $ \lim_{x \to 0} \frac{\tan^{-1} x + \ln \left( \frac{1+x}{1-x} \right) - 2x}{x^5} = \frac{2}{5} $ Statement 2: $ \lim_{x \to 1} x \left( \frac{2}{1-x} \right) = e^2 \; \text{and can be solved by the method} \lim_{x \to 1} \frac{f(x)}{g(x) - 1} $
Find the area of the region defined by the conditions: $ \left\{ (x, y): 0 \leq y \leq \sqrt{9x}, y^2 \geq 3 - 6x \right\} \text{(in square units)} $