Variance of the Sum of $3$ Independent Trials

Question

Note: I was able to solve this question while I was typing it up, but I'll just leave this here in case someone else needs it. Feel free to improve on it in the answers.

Suppose $3$ marksmen shoot at a target. The $i$th marksman fire $n_i$ times, hitting the target each time with probability $p_i$, independently of his other shots and the shots of the other marksmen. Let $X$ be the total number of shots the target is hit. Find $Var(X)$.

Let $X_i$ be the number of times the $i$th marksman hits the target.

$Var(X)$

$= Var(X_1 + X_2 + X_3)$

$= Var(X_1) + Var(X_2) + Var(X_3)$ since the shots of one marksman is independent from the others.

$= \sum_{i=1}^{3} Var(X_i)$

Let $I_i$ be the indicator that the $i$th shot hit.

$Var(X_i)$

$= Var(I_1 + I_2 + \ldots +I_{n_i})$

$= Var(I_1) + Var(I_2) + \ldots + Var(I_{n_i})$ since each shots of one marksman is independent of his other shots.

$= \sum_{j=1}^{n_i}Var(I_j)$

Now we have: $Var(X)= \sum_{i=1}^{3} \sum_{j=1}^{n_i}Var(I_j)$

$Var(I_j)$

$=E(I_j^2) - (E(I_j))^2$

$= (1^2 \cdot p_i+0^2 \cdot (1- p_i)) - p_i^2$ considering each shot has the same probability ($p_i$).

= $p_i - p^2_i$

Thus, $Var(X)= \sum_{i=1}^{3} \sum_{j=1}^{n_i}(p_i - p^2_i) = \sum_{i=1}^{3}n_i(p_i - p^2_i) = \sum_{i=1}^{3}n_ip_i(1 - p_i)=\sum_{i=1}^{3}n_ip_iq_i$

where $q_i = 1 - p_i$

Answer 1

$X_i \sim Bin(n_i, p_i)$, hence $\operatorname{Var}(X_i)=n_ip_iq_i$

Hence $$\operatorname{Var}(X)= \sum_{i=1}^3 n_ip_iq_i,$$

where $q_i=1-p_i$.

Answer 2

Yes. Well reasoned.

$$\mathsf {Var}(X) = \sum_{i=1}^3 n_i p_i(1-p_i)\qquad\checkmark$$

You could also recognise that the $X_i$ are binomially distributed counts.   (The variance of which is classically obtained by the indicator random variable method anyway.)