HELP ME PLZ!!!!!!!!probability

2013-04-16 4:13 am
A bag contains 3 black balls, 3 green balls and 3 yellow balls. Calvin repeats drawiong one ball at a time randomly from the bag without replacement until a green ball is drawn. Find the probability that he needs at most 3 draws.

答案係:
Need 1 draw: 3/9
Need 2 draw : (6/9)*(3/8)=0.25
Need 3 draw: (6/9)*(5/8)*(3/7)= 5/28
probability that he needs at most 3 draws=(3/9)+0.25+(5/28)=16/21

但係我個計法係:
Need 1 draw: (3C1)/(9C1) = 3/9
Need 2 draw: (6C1 * 3C1 ) / 9C2 = 1/2
Need 3 draw: (6C1 * 5C1 * 3C1) / 9C3 = 15/14
probability that he needs at most 3 draws=3/9 + 1/2 + 15/14 =40/21


我都知我個計法一定錯,因為probability 係唔會大過1

但係我唔明我個計法犯左咩毛病,錯左咩concept???

help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

回答 (1)

2013-04-16 11:51 pm
✔ 最佳答案
主要係個分母問題

因為係 “without replacement“ ,每攞一次波嘅總數(分母)就 -1

(3C1)/(9C1) = 3/9
e個無問題

(6C1 * 3C1 ) / 9C2 = 1/2
如果咁寫,意思就係同時系9個波入面抽出1個綠波同1個非綠波
6C1 ---> 6個非綠波中抽一個
3C1 ---> 3個綠波中抽一個
9C2 ---> 在所有波(9個)中「同時」抽出2個 (不考慮次序)

(6C1 * 5C1 * 3C1) / 9C3 = 15/14
e個一睇就知道錯,原因係:6+5+3>9
即係話你想系14個波入面抽3個(分子),但系個袋只有9個(分母)
6C1 ---> 6個波抽一個
5C1 ---> 5個波抽一個
3C1 ---> 3個波抽一個
以上全部乘曬嘅話即係,一次過系14個波入面抽3種唔同嘅波出黎
9C3 ---> 9個波入面「同時」抽出3個(不考慮次序)

主要問題只係諗錯野姐,注意番“C”嘅用法就可以啦

重有唔明可以留言,我會盡力補充(雖說我D表達能力唔太好......)

2013-04-17 10:25:18 補充:
變左14個其實係因為用左乘號

"C" 係用黎搵出現的次數(數量),而唔係機會率
如果乘曬嘅話, 就代表係同時出現嘅次數,無時間上嘅先後次序

即係話,如果3個"C"乘左,又用同一個分母嘅話,就表示「6個非綠波、5個非綠波、3個綠波」係同時間,系同一個袋入面抽嘅~
亦即係有(6+5+3) = 14個波

2013-04-17 10:54:58 補充:
機會率黎講,以 aCb 為例

如果真係要用乘,分母嘅a同b都會大過或等同分子嘅a同b

(6C1 * 5C1 * 3C1) / aCb
a就需要大於或等於(6+5+3)
b一樣大於或等於(1+1+1)

如果唔係就應該邊度有寫錯野
參考: 自己,數學老師(?), 最好問番數學老師,我都唔知點解釋好...


收錄日期: 2021-04-13 19:25:44
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