更新1:
How did you come up with the term: (m + n)! / (m! * n!) permutations? It seems like you are using mCn. But I thought we should use mPn.
A question about probability
2012-02-04 11:48 pm
suppose that each child born to a couple is likely to be aboy or a girl, independently of the sex distribution of the other children inthe family. For a couple having 5 children, compute the probability of havingexactly 3 are boys? Ans: 5/16Can you show me a mathmetical way for solving this probleminstead of listing all the permutations?
回答 (2)
2012-02-05 1:51 am
✔ 最佳答案
P(exactly 3 boys)= P(3 boys 2 girls)
= P(BBBGG , BBGBG , BBGGB , BGBBG , BGBGB ,
BGGBB , GBBBG , GBBGB , GBGBB or GGBBB )
= 5! / (3! * 2!) * (1/2)⁵
= 10 / 2⁵
= 5/16
2012-02-04 17:57:27 補充：
There are (m + n)! / (m! * n!) permutations of A , B for m A and n B.
2012-02-05 02:26:40 補充：
(m + n)! / (m! * n!) is another formula neither mPn nor mCn.
If there are m items of A and n items of B ,
then m items of A and n items of B
have (m + n)! / (m! * n!) permutations , it is an important formula.
2012-02-05 02:34:25 補充：
Yes
(m + n + ... + z)! / (m! * n! * ... * z!)
2012-02-05 10:29 am
So is this formula apply to three items or more?
收錄日期: 2021-04-21 22:23:33
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https://hk.answers.yahoo.com/question/index?qid=20120204000051KK00520