Answer
See prove below.
Work Step by Step
Let $n=3$ and $p$ be the success rate, we can calculate the binomial distribution as
$P(0)=_3C_0\cdot p^0(1-p)^3$
$P(1)=_3C_1\cdot p^1(1-p)^2$
$P(2)=_3C_2\cdot p^2(1-p)^1$
$P(3)=_3C_3\cdot p^3(1-p)^0$
We can calculate the mean as
$\mu=\sum X\cdot P(X)=0\times P(0)+1\times _3C_1\cdot p^1(1-p)^2+2\times _3C_2\cdot p^2(1-p)^1+3\times _3C_3\cdot p^3(1-p)^0=3p(1-p)^2+6p^2(1-p)+3p^3=3p(1-2p+p^2)+6p^2-6p^3+3p^3=3p$
Thus, we proved that the mean $\mu=3p$
