I don’t understand why this is the formula for variance for binomial distribution. The 1-p especially confuses me. The variance in the square of the standard deviation which I don’t get how this gives us a deviation.

  • 4
    $\begingroup$ Have you tried plugging into the definition of variance? $\endgroup$
    – saulspatz
    Jul 10, 2019 at 19:27
  • 1
    $\begingroup$ When $p=0$ or $p=1$, the distribution is deterministic and has zero variance. So the factors $p$ and $1-p$ were to be expected. $\endgroup$
    – user65203
    Jul 10, 2019 at 19:31

1 Answer 1


The expectation is given by

$$E(k)=\sum_{k=0}^n k\frac{n!}{k!(n-k)!} p^kq^{n-k}=np\sum_{k=1}^n \frac{(n-1)!}{(k-1)!(n-k)!} p^{k-1}q^{n-k}=np(p+q)^{n-1}=np$$ (the term $k=0$ vanishes).

Then the expectation of the square is

$$E(k^2)=\sum_{k=0}^n k^2\frac{n!}{k!(n-k)!} p^kq^{n-k}=\sum_{k=0}^n (k(k-1)+k)\frac{n!}{k!(n-k)!} p^kq^{n-k}\\ =n(n-1)p^2\sum_{k=2}^n \frac{(n-2)!}{(k-2)!(n-k)!} p^{k-2}q^{n-k}+np=n(n-1)p^2(p+q)^{n-2}+np$$ (the terms $k=0,1$ vanish).




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