紅白球取球模擬實驗
法一：一次取出
紅球數：0
法二：取後不放回
紅球數：0
法三：取後放回
紅球數：0
精選題目
題目：袋中有 $15$ 顆白球與 $5$ 顆紅球，隨機取出 $3$ 球，求取出紅球數量 $X$ 的期望值 $E(X)$：
(1) 一次取出 3 球
$$E(X) = 0 \cdot \frac{C^{5}_{0}C^{15}_{3}}{C^{20}_{3}} + 1 \cdot \frac{C^{5}_{1}C^{15}_{2}}{C^{20}_{3}} + 2 \cdot \frac{C^{5}_{2}C^{15}_{1}}{C^{20}_{3}} + 3 \cdot \frac{C^{5}_{3}C^{15}_{0}}{C^{20}_{3}}$$
$$= \frac{0 \cdot 455 + 1 \cdot 525 + 2 \cdot 150 + 3 \cdot 10}{1140} = \frac{855}{1140} = \mathbf{0.75}$$
(2) 逐次取出 1 球，取後不放回
考慮順序，分母為 $P^{20}_{3} = 6840$：
$$E(X) = \frac{0 \cdot (15 \cdot 14 \cdot 13) + 1 \cdot (3 \cdot 5 \cdot 15 \cdot 14) + 2 \cdot (3 \cdot 5 \cdot 4 \cdot 15) + 3 \cdot (5 \cdot 4 \cdot 3)}{6840}$$ $$= \frac{0 + 3150 + 1800 + 180}{6840} = \frac{5130}{6840} = \mathbf{0.75}$$
$$E(X) = \frac{0 \cdot (15 \cdot 14 \cdot 13) + 1 \cdot (3 \cdot 5 \cdot 15 \cdot 14) + 2 \cdot (3 \cdot 5 \cdot 4 \cdot 15) + 3 \cdot (5 \cdot 4 \cdot 3)}{6840}$$ $$= \frac{0 + 3150 + 1800 + 180}{6840} = \frac{5130}{6840} = \mathbf{0.75}$$
(3) 逐次取出 1 球，取後放回
單次取中紅球機率 $p = \frac{5}{20} = \frac{1}{4}$，白球機率 $q = \frac{3}{4}$。
各項機率使用分數表示（分母皆為 $4^3 = 64$）：
$$P(X=0) = C^{3}_{0}\left(\frac{1}{4}\right)^0\left(\frac{3}{4}\right)^3 = \frac{1 \cdot 1 \cdot 27}{64} = \frac{27}{64}$$ $$P(X=1) = C^{3}_{1}\left(\frac{1}{4}\right)^1\left(\frac{3}{4}\right)^2 = \frac{3 \cdot 1 \cdot 9}{64} = \frac{27}{64}$$ $$P(X=2) = C^{3}_{2}\left(\frac{1}{4}\right)^2\left(\frac{3}{4}\right)^1 = \frac{3 \cdot 1 \cdot 3}{64} = \frac{9}{64}$$ $$P(X=3) = C^{3}_{3}\left(\frac{1}{4}\right)^3\left(\frac{3}{4}\right)^0 = \frac{1 \cdot 1 \cdot 1}{64} = \frac{1}{64}$$
各項機率使用分數表示（分母皆為 $4^3 = 64$）：
$$P(X=0) = C^{3}_{0}\left(\frac{1}{4}\right)^0\left(\frac{3}{4}\right)^3 = \frac{1 \cdot 1 \cdot 27}{64} = \frac{27}{64}$$ $$P(X=1) = C^{3}_{1}\left(\frac{1}{4}\right)^1\left(\frac{3}{4}\right)^2 = \frac{3 \cdot 1 \cdot 9}{64} = \frac{27}{64}$$ $$P(X=2) = C^{3}_{2}\left(\frac{1}{4}\right)^2\left(\frac{3}{4}\right)^1 = \frac{3 \cdot 1 \cdot 3}{64} = \frac{9}{64}$$ $$P(X=3) = C^{3}_{3}\left(\frac{1}{4}\right)^3\left(\frac{3}{4}\right)^0 = \frac{1 \cdot 1 \cdot 1}{64} = \frac{1}{64}$$
期望值推導：
$$E(X) = 0 \cdot \left(\frac{27}{64}\right) + 1 \cdot \left(\frac{27}{64}\right) + 2 \cdot \left(\frac{9}{64}\right) + 3 \cdot \left(\frac{1}{64}\right)$$ $$= \frac{0 + 27 + 18 + 3}{64} = \frac{48}{64} = \frac{3}{4} = \mathbf{0.75}$$
$$E(X) = 0 \cdot \left(\frac{27}{64}\right) + 1 \cdot \left(\frac{27}{64}\right) + 2 \cdot \left(\frac{9}{64}\right) + 3 \cdot \left(\frac{1}{64}\right)$$ $$= \frac{0 + 27 + 18 + 3}{64} = \frac{48}{64} = \frac{3}{4} = \mathbf{0.75}$$
結論：無論取法為何，結果皆為 0.75 球。
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