devell 寫到:一袋中有4個白球,3個綠球,從袋中取球兩次,每次同時取出兩球,取出的球不放回袋中,
請問第二次同時取出兩個綠球的機率是多少?
按部就班的解法:
第一次有三種可能:
(White,White): 4/7 * 3/6 = 2/7
(White,Green): 4/7 * 3/6 + 3/7 * 4/6 = 4/7
(Green,Green): 3/7 * 2/6 = 1/7
第二次都拿綠球:
(W,W)(G,G) = 2/7 * ( 3/5 * 2/4) = 6/70
(W,G)(G,G) = 4/7 * ( 2/5 * 1/4) = 4/70
(G,G)(G,G) = 1/7 * (1/5 * 0/4) = 0
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加總: 6/70 + 4/70 = 1/7
如果白球有w顆, 綠球有g顆,
第一次有三種可能:
(W,W) = w/(w+g) * (w-1)/(w+g-1)
(W,G) = 2wg/(w+g)/(w+g-1)
(G,G) = g/(w+g) * (g-1)/(w+g-1)
第二次都拿綠球:
(W,W)(G,G) = w(w-1)/(w+g)/(w+g-1) * { g(g-1)/(w+g-2)/(w+g-3) }
(W,G)(G,G) = 2wg/(w+g)/(w+g-1) * { (g-1)(g-2)/(w+g-2)/(w+g-3) }
(G,G)(G,G) = g(g-1)/(w+g)/(w+g-1) * { (g-2)(g-3)/(w+g-2)/(w+g-3) }
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加總起來即可
分子: g(g-1)[ w(w-1) + 2w(g-2) + (g-2)(g-3)] = g(g-1)(w+g-2)(w+g-3)
分母: (w+g)(w+g-1)(w+g-2)(w+g-3)
所以最終為 g(g-1)/(w+g)/(w+g-1)
與Tzwan 的推論一致