Re: [童年] 殘念的99分
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解給你看 根據A 勝率 1/3 B 勝率 2/3
那麼用簡單機率列表 先將B拆解成B b
於是列出4場的A獲勝的樣本空間數
{AAAA , AAAB , AAAb , AABA, AAbA, ABAA, AbAA, BAAA, bAAA}9項
母體樣本數 3^4
所以 9 / 81 = 1/9
請問 答案哪裡錯了
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◆ From: 111.249.50.175
推
04/20 13:10, , 1F
04/20 13:10, 1F
用不是簡單機率的概念算一下
列出A獲勝的狀況 {AAA,AABA,ABAA, BAAA}
AAA獲勝狀況應該為 (1/3)^3 = 1/27
BAAA, ABAA, AABA 個別應該為 (2/3)*(1/3)^3 = 2/81
所以總和應該是 (3 + 2 + 2 + 2) / 81 = 9 / 81 = 1/9
請問 7/81 怎麼算
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※ 編輯: simonjen 來自: 111.249.50.175 (04/20 13:16)
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"
中學的時候也碰到一個天才老師
一直堅持"機率=特定事件數/所有事件數"
"
以上節錄原文
這就是簡單機率概念 謝謝
※ 編輯: simonjen 來自: 111.249.50.175 (04/20 13:30)
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話說很久以前有一個問題
有一隻蒼蠅在兩台相距1000KM要相撞的車子之間來回飛行且由某一車出發
蒼蠅速度為時速1300KM
兩車個別的時速為200KM 800KM
問 車子相撞前蒼蠅飛了多長距離
※ 編輯: simonjen 來自: 111.249.50.175 (04/20 13:41)
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