105指考數學甲選填第11題詳解

設 $X_1, X_2, X_3 \in \{0, 1\}$ 分別表示第 $1, 2, 3$ 次轉動後指針所停區域的數字。遊戲數字之和為 $S = X_1 + X_2 + X_3$。根據期望值的加法性質： $$E[S] = E[X_1] + E[X_2] + E[X_3]$$ 由於 $X_i \in \{0, 1\}$，故 $E[X_i] = 1 \times P(X_i = 1) + 0 \times P(X_i = 0) = P(X_i = 1)$。我們只需計算每一次轉動後停在區域 $1$ 的機率。 已知起點 $X_0 = 1$。 - 第一次轉動： $$P(X_1 = 1) = P(X_1 = X_0) = \dfrac{1}{4} \implies E[X_1] = \dfrac{1}{4}$$ - 第二次轉動： $$P(X_2 = 1) = P(X_2 = 1 \mid X_1 = 1)P(X_1 = 1) + P(X_2 = 1 \mid X_1 = 0)P(X_1 = 0) = \dfrac{1}{4} \times \dfrac{1}{4} + \dfrac{3}{4} \times \left(1 - \dfrac{1}{4}\right)$$ $$P(X_2 = 1) = \dfrac{1}{16} + \dfrac{9}{16} = \dfrac{10}{16} = \dfrac{5}{8} \implies E[X_2] = \dfrac{5}{8}$$ - 第三次轉動： $$P(X_3 = 1) = P(X_3 = 1 \mid X_2 = 1)P(X_2 = 1) + P(X_3 = 1 \mid X_2 = 0)P(X_2 = 0) = \dfrac{1}{4} \times \dfrac{5}{8} + \dfrac{3}{4} \times \left(1 - \dfrac{5}{8}\right)$$ $$P(X_3 = 1) = \dfrac{5}{32} + \dfrac{9}{32} = \dfrac{14}{32} = \dfrac{7}{16} \implies E[X_3] = \dfrac{7}{16}$$ 計算期望值之和： $$E[S] = \dfrac{1}{4} + \dfrac{5}{8} + \dfrac{7}{16} = \dfrac{4}{16} + \dfrac{10}{16} + \dfrac{7}{16} = \dfrac{21}{16}$$ 故此遊戲期望值為 $\dfrac{21}{16}$。