Correct Answer: D \(\,P(A \cap B)=\frac{1}{4}\,\) To find the statement that is not true, we calculate the probabilities for each event based on the given sample space. Step 1: Define the Sample Space \(\Omega\) The sample space \(\Omega\) consists of all ordered pairs \((x,y)\) where \(x,y\in\{1,2,3,4\}\). Total outcomes: \[ 4 \times 4 = 16. \] So the probability of any single outcome is \(\frac{1}{16}\). Step 2: Evaluate \(P(A)\) Event \(A\) is the set of outcomes where \(x \ge 2\). Hence \(x\in\{2,3,4\}\). For each of these \(x\)-values, \(y\) can be \(1,2,3,4\) (4 outcomes each). Total outcomes in \(A\): \[ 4+4+4=12. \] Thus, \[ P(A)=\frac{12}{16}=\frac{3}{4}. \] Therefore, statement B is true. Step 3: Evaluate \(P(B)\) Event \(B\) is the set of outcomes where \(y>x\). Count outcomes: \[ x=1 \Rightarrow y\in\{2,3,4\} \ (3) \] \[ x=2 \Rightarrow y\in\{3,4\} \ (2) \] \[ x=3 \Rightarrow y\in\{4\} \ (1) \] \[ x=4 \Rightarrow \text{none} \ (0) \] Total outcomes in \(B\): \[ 3+2+1=6. \] Thus, \[ P(B)=\frac{6}{16}=\frac{3}{8}. \] Therefore, statement A is true. Step 4: Evaluate \(P(A\cap B)\) Event \(A\cap B\) means \(x\ge 2\) and \(y>x\). Count outcomes: \[ x=2 \Rightarrow y\in\{3,4\} \ (2) \] \[ x=3 \Rightarrow y\in\{4\} \ (1) \] \[ x=4 \Rightarrow \text{none} \ (0) \] Total outcomes in \(A\cap B\): \[ 2+1=3. \] Thus, \[ P(A\cap B)=\frac{3}{16}. \] Statement D claims \(P(A\cap B)=\frac{1}{4}=\frac{4}{16}\), but we got \(\frac{3}{16}\). So statement D is false. Step 5: Check for Independence Events are independent if \[ P(A\cap B)=P(A)P(B). \] Compute: \[ P(A)P(B)=\frac{3}{4}\cdot\frac{3}{8}=\frac{9}{32}. \] But \[ \frac{3}{16}\neq \frac{9}{32}. \] So \(A\) and \(B\) are not independent, hence statement C is true. Conclusion Statements A, B, and C are true, while statement D is not true.