NIMCET 2025 Paper

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