Consider the sample space $\Omega={\{(x,y):x,y\in{\{1,2,3,4\}\}}}$ where each outcome is equally likely. Let A = {x ≥ 2} and B = {y > x} be two events. Then which of the following is NOT true?

We have the sample space \[ \Omega = \{(x,y) : x,y \in \{1,2,3,4\}\}, \qquad |\Omega| = 16. \] Event \[ A = \{x \ge 2\}. \] Values of \(x = 2,3,4\), so total favorable outcomes: \[ 12 \quad \Rightarrow \quad P(A) = \frac{12}{16} = \frac{3}{4}. \] Event \[ B = \{y > x\}. \] Count pairs: \[ \begin{aligned} x=1 &: (1,2),(1,3),(1,4) \Rightarrow 3, \\ x=2 &: (2,3),(2,4) \Rightarrow 2, \\ x=3 &: (3,4) \Rightarrow 1. \end{aligned} \] Thus total = 6, so \[ P(B) = \frac{6}{16} = \frac{3}{8}. \] Now compute \(A \cap B\): \[ x \ge 2,\quad y > x. \] Valid pairs: \[ (2,3),(2,4),(3,4). \] So \[ P(A \cap B) = \frac{3}{16}. \] Check independence: \[ P(A)P(B) = \frac{3}{4} \cdot \frac{3}{8} = \frac{9}{32}, \] but \[ P(A \cap B) = \frac{3}{16} = \frac{6}{32}. \] Since \[ \frac{9}{32} \neq \frac{6}{32}, \] events \(A\) and \(B\) are not independent. Therefore, the NOT true statement is: \[ \boxed{P(A \cap B) = \frac14}. \]