$\newcommand{\E}{\mathrm{E}}$This has been bothering me for a while. I have a question which gives that you have two types of drivers risky and safe, with safe being 80% of population ($p =0.8$). Now you take a $100$ drivers randomly from the population, $S$ is the number of safe and $R$ is the number of risky. It asks to find the Var($S$).
My first approach was define $X_i$ to be $1$ if the ith person is safe and $0$ if the ith person is risky. Then we can define $S = X_1 + X_2 \dots X_{100}$. It would seem that $S$ is binomial so the variance should be:
Var($S$) = $100p(1-p) = 100\times0.8\times02 = 16$.
Then I tried it using an alternative approach:
Var($S$) = $\E S^2 - (\E S)^2$. I know the $\E S = 100 \times 0.8 = 80$. Now for $\E S^2$.
$\E S^2 = \E[(X_1 + X_2 \dots X_{100})(X_1 + X_2 \dots X_{100})] = \sum_{i=1}^{100} \sum_{j=1}^{100} \E[X_i X_j] = \sum_{i=1}^{100}\sum_{j=1}^{100}0.64 = 6400$.
Then Var($S$) $= 6400 - 80^2 = 0.$ This doesn't seem right and contradicts my answer from earlier. Either the RV is not binomial or I've made some error in my variance calculation. I believe it is the latter, since a variance of $0$ doesn't make sense.
FIGURED IT OUT! CHECK COMMENTS
$\sum_{i=1}^{100} \sum_{j=1}^{100} \E[X_i X_j] = \sum_{i=1}^{100} \E[X_i^2] + \sum_{i \ne j}^{9,900}\E[X_iX_j] = 100\times0.8 + 9,900 \times 0.64 = 6,416$
Var($S$) = $6,416 - 80^2 = 16$!