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?
 A: $\newcommand{\Var}{\operatorname{Var}}$
I would like to flesh out some details to provide possible guidance.

It seems like the scenario is something like there are three archers, $A,B,C$, and each has accuracy $1/8,1/4,$ and $1/2$ respectively. Find the expectation and varaince of the number of successful shots.
Let $X_i$ for $i = A,B,C$ denote the number of successful shots. Then, yes, each individual shot is a Bernoulli trial with success $1/8,1/4$ and $1/2$ for each archer. 
However, notice that $X_A$ is the number of successes in $n = 3$ independent trials with chance $1/8$ of success. Hence 
$$X_A\sim\text{Binomial}(3, 1/8).$$
Similarly, 
$$X_B\sim\text{Bin}(5,1/4)$$
and
$$X_C\sim\text{Bin}(2, 1/2).$$
If we call $X = X_A+X_B+X_C$ the number of total successful shots, then
$$E[X] =E[X_A]+E[X_B]+E[X_C] = 3\cdot\frac{1}{8}+5\cdot\frac{1}{4}+2\cdot\frac{1}{2} = \frac{21}{8},$$ and by independence 
$$\Var(X) = \Var(X_A)+\Var(X_B)+\Var(X_C) = 3\cdot\frac{1}{8}\cdot\frac{7}{8}+5\cdot\frac{1}{4}\cdot\frac{3}{4}+2\cdot\frac{1}{2}\cdot\frac{1}{2} = \frac{113}{64}.$$
Notice that $X$ is does not follow a binomial distribution.
A: 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):
$$\begin{align}
\mathsf {Var}(X_A+X_B+X_C)
 ~=~& \mathsf {Var}(X_A)+\mathsf {Var}(X_B)+\mathsf {Var}(X_C)
\\[1ex]
 ~=~& \frac{3\cdot7}{64}+\frac{5\cdot 3}{16}+\frac{2}{4}
\\[1ex] ~=~& \frac{113}{64} & \boxed{\raise{2ex}~=\lower{1ex}~ 1.765625~}\end{align}$$

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