Six balls A, B, C, D, E and F are put into six boxes numbered 1, 2, 3, 4, 5, and 6. If only one ball can be put into each box, what is the probability that ball A is not in box 1 and ball B is not in box 6?
http://hk.knowledge.yahoo.com/question/question?qid=7007110510391
這條題目之前有人答過..但我唔明佢既計法
[4*4+1*5]*4*3*2*1係點樣諗出黎?
或者有冇易明D既方法?
F.5 math~probability(2)
2011-04-03 3:14 am
回答 (1)
2011-04-03 3:34 am
✔ 最佳答案
Just using the inclusion-exclusion principle is OK. The no. of choice= N(no restriction) - N(A is in box 1) - N(B is in box 6) + N(A and B is in box 6)
= 6! - 5! - 5! + 4!
= 504
So, P(ball A is not in box 1 and ball B is not in box 6)
= 504/720
= 7/10
Hope that helps !
收錄日期: 2021-04-13 17:53:42
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