Finding the maximum and minimum variance of the sum of two Bernoulli events? You are guessing the contents of two envelopes. Let $U_i$ be the event that you guess correctly. Your probability of guessing correctly for each envelope is $P(U_1) = P(U_2) = 3/4$. $U_1$ and $U_2$ may or may not be independent of each other. Let $X$ be a random variable that represents the total number of correct guesses you've made. $X$ can be $0$, $1$, or $2$.
a. Compute  $()$.
So I modeled this as a sum of two Bernoulli events; each one has $p = 0.75$, and can take on $1$ for success, or $0$ for failure.
$$E(X) = 0.75 + 0.75 = 1.5$$
b. Compute the maximum value for $\operatorname{var}()$.
c.* Compute the minimum value for $\operatorname{var}()$.
I'm unclear how to answer these two questions. 


*

*Now, I know $$\operatorname{var}(X) = \operatorname{var}(U_1 + U_2) = \operatorname{var}(U_1) + \operatorname{var}(U_2) + 2\operatorname{cov}(U_1,U_2)$$

*I also know cov = $E(XY)-E(X)E(Y)$

*I also know that $E(X) = p$ for a Bernoulli variable, and that $\operatorname{var}(X) = pq$ for a Bernoulli variable.


I have no idea how to calculate or rationalize an $E(XY)$ to maximize or minimize the variance. Is this in the right direction?
Does $E(XY)$ have to take on a value of either $1$ or $0$?
Now suppose there are four envelopes, $1$ through $4$. Again, let  $_$  be the event that you guess envelope i correctly.  $(U_i)=3/4$  for all  $$ . Once again, let random variable  $$  be the total number of envelopes that are correctly guessed.
d.* Compute the minimum and maximum values for $\operatorname{var}()$.
I likewise would have no idea how to do this. Following the above line of thinking, cov() with multiple variables would give a matrix. How would this be incorporated?
 A: You're correct with (a). We always have the following inequality for the covariance:
$$|\operatorname{cov}(X,Y)|\leq \sqrt{\operatorname{var}(X)\operatorname{var}(Y)}$$
This means we have $|\operatorname{cov}(U_1,U_2)|\leq pq$, so $0\leq\operatorname{var}(U_1+U_2)\leq 4pq$
The upper bound occurs when $U_1=U_2$. This can give you an idea what happens in four variable case, and maximum when one equals the other three (i.e. $16pq$). 
We can't reach this lower bound because $U_1+U_2$ cannot be made constant in any circumstance (as in @whuber's comment). Here, Consider making a 2x2 table of $U_1,U_2$, where $P(U_1=1,U_2=1)=x$. Then, $P(U_1=1,U_2=0)=P(U_1=0,U_2=1)=3/4-x$, and $P(U_1=0,U_2=0)=x-1/2$. The covariance is $E[U_1U_2]-E[U_1]E[U_2]$. It is minimized when $E[U_1U_2]=P(U_1=1,U_2=1)=x$ is minimum. For each probability be non-negative, we need $1/2\leq x\leq 3/4$, which yield $\min[\operatorname{cov}(U_1,U_2)]=1/2-9/16=-1/16$, and therefore minimum variance will be $1/4$.
For four variables, the variance can be $0$ (i.e. minimum) when the total is constant, i.e. $U_1+U_2+U_3+U_4=3$, which doesn't contradict with the expectations.
