標題:
probability
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發問:
There are 3 bags X,Y and Z. Bag X contains 1 red card and 3 green cards, bag Y contains 8 red cards and 4 green cards, and bag Z contains 2 red cards and 4 green cards. A card is drawn at ramdom from each of the 3 bags. It is known that 2 red cards and 1 green card care drawn. If a card is now drawn at random from... 顯示更多 There are 3 bags X,Y and Z. Bag X contains 1 red card and 3 green cards, bag Y contains 8 red cards and 4 green cards, and bag Z contains 2 red cards and 4 green cards. A card is drawn at ramdom from each of the 3 bags. It is known that 2 red cards and 1 green card care drawn. If a card is now drawn at random from bag Z, find the probability that it is red. 更新: Ans = 3/11, but do not know why? 更新 2: What is 56/88 What is 48+8+32 how do we get 7/11 and 4/11 ? 更新 3: how to get (1/5)(7/11) + (2/5)(4/11) ?
最佳解答:
P(2 red cards and 1 green card) = P(X green) + P(Y green) + P(z green) = (3/4)(8/12)(2/6) + (1/4)(4/12)(2/6) + (1/4)(8/12)(4/6) = 11/36 56/88 48+8+32 P(the probability that the second card drawn from Z is red) = P(second card is red | first card of Z is red) + P(second card is red | first card of Z is green) = (1/5)(7/11) + (2/5)(4/11) = 15/55 = 3/11 2011-03-18 13:38:19 補充： Please omit 56/88 48+8+32
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