1. Sample space and basic probabilities

When two fair six-faced dice $A$ and $B$ are thrown simultaneously, the total number of equally likely outcomes is $6\times 6=36.$

We define:

• $E_1$: die $A$ shows $4$
• $E_2$: die $B$ shows $2$
• $E_3$: sum on the two dice is odd

So, $P(E_1)=\frac{6}{36}=\frac{1}{6},$ because if die $A=4$, die $B$ can be anything.

Similarly, $P(E_2)=\frac{6}{36}=\frac{1}{6}.$

Now for $E_3$: The sum is odd when one die is odd and the other is even. Each die has $3$ odd and $3$ even faces, so P(E_3)=\frac{3\cdot 3+3\cdot 3}{36}=rac{18}{36}=rac{1}{2}.

2. Check option A: $E_1$ and $E_2$ are independent

We compute: $P(E_1\cap E_2)=\frac{1}{36},$ since this is the single outcome $(4,2)$.

Also, $P(E_1)P(E_2)=\frac{1}{6}\cdot \frac{1}{6}=\frac{1}{36}.$

Thus, $P(E_1\cap E_2)=P(E_1)P(E_2),$ so $E_1$ and $E_2$ are independent.

Hence, A is true.

3. Check option B: $E_2$ and $E_3$ are independent

If $E_2$ occurs, then die $B=2$ (which is even). For the sum to be odd, die $A$ must be odd: $1,3,5$. So favorable outcomes are: $(1,2),(3,2),(5,2),$ which are $3$ outcomes.

Therefore, $P(E_2\cap E_3)=\frac{3}{36}=\frac{1}{12}.$

Now, $P(E_2)P(E_3)=\frac{1}{6}\cdot \frac{1}{2}=\frac{1}{12}.$

Thus, $P(E_2\cap E_3)=P(E_2)P(E_3),$ so $E_2$ and $E_3$ are independent.

Hence, B is true.

4. Check option C: $E_1$ and $E_3$ are independent

If $E_1$ occurs, then die $A=4$ (which is even). For the sum to be odd, die $B$ must be odd: $1,3,5$. So favorable outcomes are: $(4,1),(4,3),(4,5),$ which are again $3$ outcomes.

Therefore, $P(E_1\cap E_3)=\frac{3}{36}=\frac{1}{12}.$

And, $P(E_1)P(E_3)=\frac{1}{6}\cdot \frac{1}{2}=\frac{1}{12}.$

Thus, $P(E_1\cap E_3)=P(E_1)P(E_3),$ so $E_1$ and $E_3$ are independent.

Hence, C is true.

5. Check option D: $E_1,E_2,E_3$ are independent

For three events to be independent, we need:

• pairwise independence, and

$P(E_1\cap E_2\cap E_3)=P(E_1)P(E_2)P(E_3).$

We already found pairwise independence above.

Now check the triple intersection:

• $E_1$ means $A=4$
• $E_2$ means $B=2$ So together this gives the outcome $(4,2)$ only. But then the sum is $4+2=6,$ which is even, not odd. Hence, $E_1\cap E_2\cap E_3=\varnothing,$ and therefore $P(E_1\cap E_2\cap E_3)=0.$

But $P(E_1)P(E_2)P(E_3)=\frac{1}{6}\cdot \frac{1}{6}\cdot \frac{1}{2}=\frac{1}{72}\neq 0.$

So the three events are not mutually independent.

Hence, D is false.

6. Conclusion

The statement which is NOT true is: $\boxed{\text{D}}$