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統計的二項分配問題詳細解釋(英文)~已知答案

Q: 

a coin is tossed 6 times . find the probability that the number of heads in the first 3 throws is the same as the number of heads in the last three throws .

A:

the number of heads in the first 3 throws , say X , is binomial with n = 3 and p = 1/2 .

then P〔X = 0〕= 1/8 , P〔X = 1〕= 3/8 , P〔X = 2〕= 3/8 , P〔X = 3〕= 1/8 .

if Y is the number of heads in the last 3 throws , then Y has the same distribution as X .

then , P〔X = Y〕= , P〔X = Y = 0〕+ … + , P〔X = Y = 3〕= (1/8)(1/8) + (3/8)(3/8) + (3/8)(3/8) + (1/8)(1/8) = 20/64 = 5/16 .

【(1/8)(1/8) 】這個是1/8的平方,可是我不會打平方,所以都用兩個數相乘來表示

★我也搞不懂為啥【 P〔X = Y〕= , P〔X = Y = 0〕+ … + , P〔X = Y = 3〕】這段的算式要平方呢??

沒頭沒腦的是因為【 P〔X = Y = 0〕】這裡的X=Y嗎??解釋一下吧

我的問題最主要有三個(不是說我只要問這三個,還是必需要全部都講解!!只是我最不能理解的就是這些,這三個部分ㄧ定要解釋到..)

1.題目到底是在問啥呢??我看的懂個別的單字,可是連在ㄧ起,我就完全搞不懂了...有哪些已知數,以及要求什麼??

the number of heads in【 the first 3 throws is the same as the number of heads in the last three throws】.

尤其我框起來的那段英文,我根本完全看不懂,有人可以解釋一下嗎??

2.P〔X = Y〕= , P〔X = Y = 0〕+ … + , P〔X = Y = 3〕= (1/8)(1/8) + (3/8)(3/8) + (3/8)(3/8) + (1/8)(1/8) = 20/64 = 5/16 .

這段是如何算出來的??是有公式嗎??還是其他原因??

3.【 if Y is the number of heads in the last 3 throws , then Y has the same distribution as X 】. 解答的這段文字是代表啥意思??

2 Answers

Rating
  • 1 decade ago
    Favorite Answer

    Q: a coin is tossed 6 times . find the probability that the number of heads in the first 3 throws is the same as the number of heads in the last three throws .

    問題:投擲硬幣六次,計算前三次投擲出現人頭次數與後三次出現人頭次數相等之機率

    解答: the number of heads in the first 3 throws , say X , is binomial with n = 3 and p = 1/2 .

    前三次投擲出現人頭之次數假設為X,投擲次數=3,每次投擲出現人頭之機率P=1/2

    then

    P〔X = 0〕= 1/8-前三次投擲都沒出現人頭之機率

    P〔X = 1〕= 3/8-前三次投擲僅出現一次人頭之機率

    P〔X = 2〕= 3/8-前三次投擲三次出現兩次人頭之機率

    P〔X = 3〕= 1/8-前三次投擲三次出現三次人頭之機率

    if Y is the number of heads in the last 3 throws , then Y has the same distribution as X .

    後三次投擲出現人頭之次數假設為Y,Y的條件與前述X的條件相同

    所以

    P〔Y = 0〕= 1/8-後三次投擲都沒出現人頭之機率

    P〔Y= 1〕= 3/8-後三次投擲僅出現一次人頭之機率

    P〔Y = 2〕= 3/8-後三次投擲三次出現兩次人頭之機率

    P〔Y = 3〕= 1/8-後三次投擲三次出現三次人頭之機率

    then ,求 P〔X = Y〕之機率

    因為兩個條件必須同時成立所以機率必須用乘法計算

    P〔X = Y = 0〕--前三次與後三次投擲皆未出現人頭 (1/8)(1/8)=1/64

    P〔X = Y = 1〕--前三次與後三次投擲皆出現一次人頭

    (3/8)(3/8)=9/64

    P〔X = Y = 2〕前三次與後三次投擲皆出現兩次人頭

    (3/8)(3/8)=9/64

    P〔X = Y = 3〕前三次與後三次投擲皆出現三次人頭

    (3/8)(3/8)=9/64

    前三次投擲出現人頭之次數與後三次投擲出現人頭之次數相同之情況共有如上四種(0、1、2、3),由於四種情形不可能同時發生,所以將四種情形發生之機率相加

    (1/8)(1/8) (3/8)(3/8) (3/8)(3/8) (1/8)(1/8) = 20/64 = 5/16 .

    輸入平方之方法:

    先將數字輸入,再以滑鼠左鍵點螢幕上方格式之功能鍵,選擇字型之功能,再選擇上標,案確定後游標就會跑到數字之右上方,此時,輸入2即是平方輸入3即是立方,輸入完畢後,再依上述步驟取消上標功能,不然游標不會恢復原狀喔!

    2007-12-31 12:37:20 補充:

    應該已經夠詳細了

    如過還有不懂再連絡

    2008-01-05 07:35:33 補充:

    修正

    (1/8)(1/8)+(3/8)(3/8)+(3/8)(3/8)+(1/8)(1/8) = 20/64 = 5/16

    2008-01-05 07:37:38 補充:

    修正

    P〔X = Y = 3〕前三次與後三次投擲皆出現三次人頭

    (1/8)(1/8)=1/64

    Source(s): 自己, 自己, 自己
  • 呆子
    Lv 7
    1 decade ago

    1.硬幣有人頭與非人頭兩面,前面3次投擲與後面3次投擲人頭出現機率次數是相同

    2.因X = Y就是第一項所產生結果

    3.與第一項意思相同*****

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