How do I find the Bernoulli Variance for three players shooting at a target with different probabilities of success? Given $i=\{A,B,C\}$
$A: n=3, P(\text{Success})=1/8$
$B: n=5, P(\text{Success})=1/4$
$C: n=2, P(\text{Success})=1/2$
And assuming that trials are independent for all $n$, I am trying to find the Variance of the number of times a target will be hit. 
To me this seems like a Bernoulli Random Variable, so I assume you can calculate the Variance of hits for all trials as the sum of all variances for each player. 
That gives: 
$$\text{Var}(\text{Hit})=3(1/8)(7/8)+5(1/4)(3/4)+2(1/2)(1/2).$$
Can I calculate the variance of the number of times the target will be hit in this way?

The count of total target hits is itself not a binomially distributed random variable.   It is however, the sum of three mutually independent binomial random variables: $$X_A~\sim~\mathcal{Bin}(3,1/8),\\ X_B~\sim~\mathcal{Bin}(5,1/4),\\ X_C~\sim~\mathcal{Bin}(2,1/2)$$
So, the variance of the sum is the sum of the variance (because of the mutual independence):
$$\begin{align}
\mathsf {Var}(X_A+X_B+X_C)
 ~=~& \mathsf {Var}(X_A)+\mathsf {Var}(X_B)+\mathsf {Var}(X_C)
\\[1ex]
 ~=~& \frac{3\cdot7}{64}+\frac{5\cdot 3}{16}+\frac{2}{4}
\\[1ex] ~=~& \frac{113}{64} & \boxed{\raise{2ex}~=\lower{1ex}~ 1.765625~}\end{align}$$

$\because~~Y~\sim~\mathcal {Bin}(n,p) ~\iff~ \mathsf {Var}(Y)~=~np(1-p)$
