Variance of Bernoulli Random Variable with a Random Variable as parameter Let X be a Bernoulli Random Variable whose parameter is a Uniform Random Variable Q which takes values in the domain [0 , 0.1].
We want to find var(X).  My reasoning is the following:
Using the Law of Total Variance we have:
var(X) = E[var(X|Q)] + var(E[X|Q]) = E[Q * (1 - Q)] + var(Q) = E[Q] - E[Q^2] + var(Q) = E[Q] - (var(Q) + (E[Q])^2) + var(Q) = E[Q] - (E[Q])^2 = 0.05 - 0.05^2 = 0.0475.
Do you agree with this line of reasoning?
 A: You are using the correct approach with the total variance law. 
$$\mathbb{V}ar(X)= \mathbb{V}ar(\mathbb{E}(X|Q)) + \mathbb{E}(\mathbb{V}ar(X|Q) ).$$ 
Here you will need the expectations and variances of the Bernoulli and the continuous uniform. These are: 
$$\mathbb{E}(X|Q)=Q,$$
$$\mathbb{V}ar(X|Q)=Q(1-Q),$$
$$\mathbb{E}(Q)=\frac{a+b}{2},$$
$$\mathbb{V}ar(Q)=\frac{(a-b)^2}{12},$$
where here $a=0.1$ and $b=0.0$, the limits of the continuous uniform. The only other thing I used to solve this was the variance relation
$$ \mathbb{V}ar(Y) = \mathbb{E}(Y^2)-(\mathbb{E}(Y))^2.$$
This is rearranged to get a relationship between the mean and the variance to get the second moment if you know the mean and the variance. So from the total variance law
$$\mathbb{V}ar(X)= \mathbb{V}ar(\mathbb{E}(X|Q)) + \mathbb{E}(\mathbb{V}ar(X|Q) )
=\mathbb{V}ar(Q) + \mathbb{E}(Q(1-Q))
=\mathbb{V}ar(Q) + \mathbb{E}(Q)-\mathbb{E}(Q^2)
=\frac{0.1^2}{12} + 0.05 - (\frac{.01^2}{12} + 0.05^2)
=0.0008333333 + 0.05 - (0.000833333 + 0.0025)
=0.05 - 0.0025 
=0.0475
$$ 
which matches what you have in the posting.
