Events are independent or not independent in the following example? Question. 
Let $X_1,X_2,X_3$    denote the outcomes of three rolls of a six-sided die. (I.e., each $X_i$   is uniformly distributed among ${1,2,3,4,5,6}$ and by assumption they are independent.) Let $Y$ denote the product of $X_1$ and $X_2$  and $Z$ the product of $X_2$ and $X_3$. Which of the following statements is correct?
option 1. $Y$ and $Z$ are not independent, but $E[Y*Z]=E[Y]*E[Z]$. 
option 2. $Y$ and $Z$ are independent, and $E[Y*Z]=E[Y]*E[Z]$.
option 3. $Y$ and $Z$  are independent, but $E[Y*Z] \neq E[Y]*E[Z]$.
option 4. $Y$ and $Z$  are not independent, and $E[Y*Z] \neq E[Y]*E[Z]$.

My Solution:
$X_1,X_2,X_3$ are variables each taking values   ${1,2,3,4,5,6}$ and each value has a probability $\frac{1}{6}$. 
Now,  
$Y = X_1 * X_2$ 
$Z = X_2 * X_3$
To find: $E[Y*Z] = E[Y]*E[Z]$  OR $E[Y*Z] \neq E[Y]*E[Z]$
LHS: $E[Y*Z] \\
= E[X_1 * X_2 * X_2 * X_3 ] \\
= E[X_1 * X_2 * X_3 ]    \text{(As in an event $X_2$ can have value  [1,6] only for both the $X_2$  variable above )} \\
= E[X_1] * E[X_2] * E[X_3] \text{(As $X_1$ , $X_2$ and $X_3$ are independent )} $
RHS: $E[Y]*E[Z]  \\
=  E[X_1 * X_2] * E[X_2 * X_3 ] \\
=  E[X_1] * E[X_2] * E[X_2] * E[X_3]   \text{(As $X_1$ and $X_2$ are independent and $X_2$ and $X_3$ are independent )}$ 
∴ $ LHS  \neq RHS $
option 4. $Y$ and $Z$  are not independent, and $E[Y*Z] \neq E[Y]*E[Z]$.
Can someone please tell me if my solution is valid? I know that the answer is surely option 4 but I wanted to know the correct way of proving this problem.
 A: Each of $Y$ and $Z$ could potentially take any of the values $1,2,3,4,5,6,8,10,12,15,16,\dots,36$.  The full list isn't important, but what is important is that $Y$ could potentially take values $1,2,3,4,5$ with positive probability.
Now... If we know that $Z=36$, this directly implies that the results of both $X_2$ and $X_3$ must have been $6$.  Knowing that $Z=36$, we can conclude then that $Y$ must be at least equal to $6$ or greater and that it is impossible for $Y$ to be equal to any of $1,2,3,4,5$.
Therefore, $0=Pr(Y=1\mid Z=36)\neq Pr(Y=1)>0$ and so we know that $Y$ and $Z$ are not independent.  Minimal calculation required here.

Now... $E[Y\times Z] = E[X_1\times X_2\times X_2\times X_3] = E[X_1\times X_2^2\times X_3]$
You should be able to reason that since $X_1,X_2,X_3$ are mutually independent random variables that $X_1,X_2^2,X_3$ should also be mutually independent random variables.
We get then since the expectation of a product of independent random variables is equal to the product of the expectations of independent random variables that $E[Y\times Z] = E[X_1]\times E[X_2^2]\times E[X_3]$
Compare this to $E[Y]\times E[Z] = E[X_1]\times E[X_2]\times E[X_2]\times E[X_3]$
Since $E[X_1]$ and $E[X_3]$ are clearly nonzero, these would be equal in the case that $E[X_2^2]=E[X_2]^2$.
But... We remember that $Var(X_2) = E[X_2^2]-E[X_2]^2$ and a zero variance would imply the random variable were constant.  We can clearly see that $X_2$ is not a constant random variable and so we know that $Var(X_2)$ is nonzero as well.  As such we can conclude that $E[Y\times Z]\neq E[Y]\times E[Z]$.  Again, no actual full lengthy calculations needed here.

We conclude then that option 4 is the correct answer, that $Y$ and $Z$ are not independent and that $E[Y\times Z]\neq E[Y]\times E[Z]$.
You arrived at the correct answer but by using incorrect calculations.  As mentioned in the comments above, you tried to use the equality $E[X_1\times X_2\times X_2\times X_3] = E[X_1\times X_2\times X_3]$ which simply isn't true in this and many other cases.
