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).