题目
掷六颗骰子,利用切比雪夫不等式估计六颗骰子出现点数和在15到27之间的概率为()A. geq (1)/(2)B. leq (35)/(72)C. geq (37)/(72)D. leq (37)/(72)
掷六颗骰子,利用切比雪夫不等式估计六颗骰子出现点数和在15到27之间的概率为()
A. $\geq \frac{1}{2}$
B. $\leq \frac{35}{72}$
C. $\geq \frac{37}{72}$
D. $\leq \frac{37}{72}$
题目解答
答案
C. $\geq \frac{37}{72}$
解析
步骤 1:计算单个骰子的期望和方差
单个骰子的点数可以是1到6,每个点数出现的概率是$\frac{1}{6}$。因此,单个骰子的期望值$E(X)$为:
\[ E(X) = \sum_{i=1}^{6} i \cdot \frac{1}{6} = \frac{1}{6} \sum_{i=1}^{6} i = \frac{1}{6} \cdot \frac{6 \cdot (6 + 1)}{2} = \frac{7}{2} = 3.5 \]
单个骰子的方差$Var(X)$为:
\[ Var(X) = E(X^2) - [E(X)]^2 = \sum_{i=1}^{6} i^2 \cdot \frac{1}{6} - (3.5)^2 = \frac{1}{6} \sum_{i=1}^{6} i^2 - 12.25 = \frac{1}{6} \cdot \frac{6 \cdot (6 + 1) \cdot (2 \cdot 6 + 1)}{6} - 12.25 = \frac{91}{6} - 12.25 = \frac{35}{12} \]
步骤 2:计算六颗骰子总和的期望和方差
六颗骰子的总和$S$的期望值$E(S)$为:
\[ E(S) = 6 \cdot E(X) = 6 \cdot 3.5 = 21 \]
六颗骰子的总和$S$的方差$Var(S)$为:
\[ Var(S) = 6 \cdot Var(X) = 6 \cdot \frac{35}{12} = \frac{35}{2} \]
步骤 3:应用切比雪夫不等式
根据切比雪夫不等式,对于任意随机变量$S$,有:
\[ P(|S - E(S)| \ge k) \le \frac{Var(S)}{k^2} \]
取$k = 6$,则:
\[ P(|S - 21| \ge 6) \le \frac{\frac{35}{2}}{36} = \frac{35}{72} \]
因此,六颗骰子出现点数和在15到27之间的概率为:
\[ P(15 \le S \le 27) = P(|S - 21| \le 6) \ge 1 - \frac{35}{72} = \frac{37}{72} \]
单个骰子的点数可以是1到6,每个点数出现的概率是$\frac{1}{6}$。因此,单个骰子的期望值$E(X)$为:
\[ E(X) = \sum_{i=1}^{6} i \cdot \frac{1}{6} = \frac{1}{6} \sum_{i=1}^{6} i = \frac{1}{6} \cdot \frac{6 \cdot (6 + 1)}{2} = \frac{7}{2} = 3.5 \]
单个骰子的方差$Var(X)$为:
\[ Var(X) = E(X^2) - [E(X)]^2 = \sum_{i=1}^{6} i^2 \cdot \frac{1}{6} - (3.5)^2 = \frac{1}{6} \sum_{i=1}^{6} i^2 - 12.25 = \frac{1}{6} \cdot \frac{6 \cdot (6 + 1) \cdot (2 \cdot 6 + 1)}{6} - 12.25 = \frac{91}{6} - 12.25 = \frac{35}{12} \]
步骤 2:计算六颗骰子总和的期望和方差
六颗骰子的总和$S$的期望值$E(S)$为:
\[ E(S) = 6 \cdot E(X) = 6 \cdot 3.5 = 21 \]
六颗骰子的总和$S$的方差$Var(S)$为:
\[ Var(S) = 6 \cdot Var(X) = 6 \cdot \frac{35}{12} = \frac{35}{2} \]
步骤 3:应用切比雪夫不等式
根据切比雪夫不等式,对于任意随机变量$S$,有:
\[ P(|S - E(S)| \ge k) \le \frac{Var(S)}{k^2} \]
取$k = 6$,则:
\[ P(|S - 21| \ge 6) \le \frac{\frac{35}{2}}{36} = \frac{35}{72} \]
因此,六颗骰子出现点数和在15到27之间的概率为:
\[ P(15 \le S \le 27) = P(|S - 21| \le 6) \ge 1 - \frac{35}{72} = \frac{37}{72} \]