Why is the variance of a binomial distribution not $n^2p(1-p)$? Using properties of variance, we know that $Var(aX)$ is $a^2Var(X)$.
A binomial distribution has n many Bernoulli trials, i.e. we can substitute $Var(X)$ with $Var(nB)$ (where $X$ is a binomial variable and $B$ a Bernoulli trial) which, using the property described above, gives us $n^2Var(B)$.
But, this would give us a result of $n^2p(1-p)$ rather than $np(1-p)$ which is the supposed variance of a binomial distribution.
Where is the mathematical error in the approach I have taken?
 A: It might be worth examining the binomial as a sum of $n$ i.i.d. bernoulli trials.  Let $X_i$ be i.i.d. bernoulli draws.  $Y = \sum_i X_i$ is then a binomial random variable.  The variance of this is
$$ \operatorname{Var}(Y) = \operatorname{Var}(\sum_i X_i) = \sum_i \operatorname{Var}(X_i) $$
Where I have used the property that the variances add iff the the covariance is 0 (which is true by assumption).  For a given Bernoulli random variable $\operatorname{Var}(X_i) = p(1-p)$.  Since all the $X_i$ are identical,
$$ \operatorname{Var}(Y) = \sum_i p(1-p) = n p (1-p) $$
A: The problem with your solution is at this step. $Var(X) = Var(nB)$
Because $X \neq nB$
I mean yes, $X = B_1 + B_2 + ... +B_n$ because all the $B_i$'s are $0$ or $1$, and X is the number of $1$'s in n trails. But you can't call it $nB$ because all $B$'s don't have the same value. Some of them are $0$ and some are $1$.
Your idea is only true if $B_1 = B_2 = ... = B_n = B$.
But if you really want to express binomial distribution in terms of Bernoulli's trails, you can do it like this.
$$X = B_1 + B_2 + ... +B_n$$
$$Var(X) = Var(\sum_{i=1}^n{B_i})$$
You can bring the summation out because by definition of binomial distribution, each trail is independent of the other trails. So it's like they are all independent.
$$Var(X) = \sum_{i=1}^n{Var(B_i)}$$
$$Var(X) = \sum_{i=1}^n{pq}$$
Now this you can write as $npq$ because all the $pq$'s regardless of $i$ have the same value $p(1-p)$.
$$Var(X) = npq$$
