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:
3×4=12
P(A)=‌

12

16

=‌

3

4

So statement (2) is true.
2. Compute P(B)
B={y>x}.
Count valid pairs:
- If x=1:y=2,3,4⟶3 outcomes
- If x=2:y=3,4⟶2 outcomes
- If x=3:y=4⟶1 outcome
- If x=4 : no y>x⟶0 outcomes
Total =3+2+1=6
P(B)=‌

6

16

=‌

3

8

So statement (4) is true.
3. Compute P(A∩B)
We need x≥2 and y>x.
Use values:
- x=2:y=3,4⟶2 outcomes
- x=3:y=4⟶1 outcome
- x=4 : no y>x⟶0 outcomes
Total =2+1=3
P(A∩B)=‌

3

16

But option (1) says P(A∩B)=‌

1

4

=‌

4

16

which is incorrect.
So statement (1) is NOT true.
4. Check independence
Check if
P(A∩B)=P(A)⋅P(B)
Left side:
P(A∩B)=‌

3

16

Right side:
P(A)⋅P(B)=‌

3

4

⋅‌

3

8

=‌

9

32

Compare:
‌

3

16

≠‌

9

32

So A and B are not independent.
Thus statement (3) is true.
Final Answer:
The NOT true statement is (1).