Solution
= Solution
With $\mu=\mathbb EX$,
$$
\operatorname{Var}X=\sum_x\mathbb P(X=x)(x-\mu)^2.
$$
Every summand is nonnegative. If the sum is zero, every value of positive probability equals $\mu$, so \b[$\mathbb P(X=\mu)=1$].
= Solution
With $\mu=\mathbb EX$,
$$
\operatorname{Var}X=\sum_x\mathbb P(X=x)(x-\mu)^2.
$$
Every summand is nonnegative. If the sum is zero, every value of positive probability equals $\mu$, so \b[$\mathbb P(X=\mu)=1$].