一个箱子里有$5$个相同的球，分别以$1\sim 5$标号，从中有放回地取三次，记至少取出一次的球的个数$X$，则数学期望$E(X)=$                  ．

$X$的可能取值为$1$，$2$，$3$，

$P(X=1)=\left(\dfrac{1}{5}\right)^{3}\times 5=\dfrac{1}{25}$，

$P(X=2)={\rm C}_{3}^{2}\left(\dfrac{1}{5}\right)^{2}\times \dfrac{1}{5}\times 2\cdot {\rm C}_{5}^{2}=\dfrac{12}{25}$，

$P(X=3)=\left(\dfrac{1}{5}\right)^{3}\times 6\times {\rm C}_{5}^{3}=\dfrac{12}{25}$，

$E(X)=\dfrac{1}{25}+\dfrac{24}{25}+\dfrac{36}{25}=\dfrac{61}{25}$．

故答案为：$\dfrac{61}{25}$