Why a Binomial Count Has Mean np: Indicator Variables and Variation

A count can be written as a sum of indicators: each trial contributes one for success or zero for failure. This gives a direct reason for the binomial mean formula.

Add the expected contributions

Let $I_i$ indicate success on trial $i$. Then

$$X=\sum_{i=1}^{n}I_i,\qquad E(I_i)=p.$$

Linearity of expectation gives

$$E(X)=\sum_{i=1}^{n}E(I_i)=np.$$

Independence is not needed for this expectation identity. It is needed, along with a common success probability and fixed trial count, for the ordinary binomial distribution and its familiar variance formula.

Mean and variability are different

For 20 independent trials with success probability 0.30,

$$E(X)=6,\qquad \operatorname{Var}(X)=20(0.30)(0.70)=4.2,$$

so the standard deviation is $\sqrt{4.2}\approx2.05$ successes. An expectation of six does not require exactly six successes in any one set of trials.

Know when np no longer describes a binomial model

If success probabilities differ by trial, expected contributions still add. Three indicators with probabilities 0.2, 0.4, and 0.8 have expected sum 1.4. Their sum is not an ordinary binomial count with a common success probability, even when the indicators are independent.

If the count stops when the first success occurs, the number of trials is no longer fixed. That is a different model and should not be forced into a fixed-n binomial calculation.