NIMCET 2025 Paper

1. Compute $P(A)$$A = \{x=2,3,4\}.$For each x , y can be 1-4 (4 values).So number of outcomes in A:$3 \times 4 = 12$$P(A) = \frac{12}{16} = \frac{3}{4}$So statement (2) is true.2. Compute $P(B)$$B = \{y > x\}.$Count valid pairs:- If $x=1: y=2,3,4 \longrightarrow 3$ outcomes- If $x=2: y=3,4 \longrightarrow 2$ outcomes- If $x=3: y=4 \longrightarrow 1$ outcome- If $x=4$ : no $y > x \longrightarrow 0$ outcomesTotal $=3+2+1=6$$P(B) = \frac{6}{16} = \frac{3}{8}$So statement (4) is true. 3. Compute $P(A \cap B)$We need $x \ge 2$ and $y > x$.Use values:- $x=2: y=3,4 \longrightarrow 2$ outcomes- $x=3: y=4 \longrightarrow 1$ outcome- $x=4$ : no $y > x \longrightarrow 0$ outcomesTotal $=2+1=3$$P(A \cap B) = \frac{3}{16}$But option (1) says $P(A \cap B) = \frac{1}{4} = \frac{4}{16}$ which is incorrect.So statement (1) is NOT true.4. Check independenceCheck if$P(A \cap B) = P(A) \cdot P(B)$Left side:$P(A \cap B) = \frac{3}{16}$Right side:$P(A) \cdot P(B) = \frac{3}{4} \cdot \frac{3}{8} = \frac{9}{32}$Compare:$\frac{3}{16} \neq \frac{9}{32}$So $A$ and $B$ are not independent.Thus statement (3) is true.Final Answer:The NOT true statement is (1).