Simone
【答案】
$X$的分布列为$X$$0$$1$$2$$3$$P$$\frac{1}{56}$$\frac{15}{56}$$\frac{15}{28}$$\frac{5}{28}$数学期望
$E\left( X \right)=\frac{15}{8}$．
$\frac{5}{7}$．
【解析】
$X$可取$0$，$1$，$2$，$3$，则
$P\left( x=0 \right)=\frac{\text{C}_{3}^{3}}{\text{C}_{8}^{3}}=\frac{1}{56}$，$P\left( x=1 \right)=\frac{\text{C}_{5}^{1}\text{C}_{3}^{2}}{\text{C}_{8}^{3}}=\frac{15}{56}$，$P\left( x=2 \right)=\frac{\text{C}_{5}^{2}\text{C}_{3}^{1}}{\text{C}_{8}^{3}}=\frac{30}{56}=\frac{15}{28}$，$P\left( x=3 \right)=\frac{\text{C}_{5}^{3}}{\text{C}_{8}^{3}}=\frac{10}{56}=\frac{5}{28}$，$X$的分布列为$X$$0$$1$$2$$3$$P$$\frac{1}{56}$$\frac{15}{56}$$\frac{15}{28}$$\frac{5}{28}$$E\left( X \right)=0\times \frac{1}{56}+1\times \frac{15}{56}+2\times \frac{15}{28}+3\times \frac{5}{28}=\frac{15}{8}$．
甲同学能晋级的概率
$P=P\left( X=2 \right)+P\left( X=3 \right)=\frac{15}{28}+\frac{5}{28}=\frac{5}{7}$．
