Question
Can the mean of a binomial distribution be less than its variance$?$

Answer

Let $X$ be a binomial veriate with parameters $n$ and $p.$
Mean $= np$ varience $= npq$
Mean $-$ variance $= np - npq = np (1 - q) = np.p = np^2$ 
Mean $-$ variance $ >0$ Mean $ > $ variance
So, mean can never be less than variance.

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