b*******t
发帖数: 390 | 1 有一个5×5共25个方格的盘子，往里面扔小皮球，因为小皮球比每个格子小一点，而且
弹性好，容易滚动，所以假设每次都有一个小皮球能进入到一个格子中（也不可能一个
格子出现多个球），
共有12个小皮球，25个方格中，任意五个球在一条直线或斜线上，就中奖了！
请问：中奖的概率有多大？ | w******o
发帖数: 578 | 2 (12*nchoosek(20,7)-nchoosek(12,2)*nchoosek(15,2))/nchoosek(25,12)=0.1775 | h********y
发帖数: 540 | 3 我的理解是，这个方格子总共12条线，每条线有5个格子，如果有5个球在一条线上，那
么有12种可能，而另外7个球会在剩下的20个方格里的任何方格，会有（20，7）个组合
。12个球在25个格子上的组合是（25，12），那么中奖概率就是12*（20，7）/（25，
12）=0.17888，excel算的
【在 b*******t 的大作中提到】
: 有一个5×5共25个方格的盘子，往里面扔小皮球，因为小皮球比每个格子小一点，而且
: 弹性好，容易滚动，所以假设每次都有一个小皮球能进入到一个格子中（也不可能一个
: 格子出现多个球），
: 共有12个小皮球，25个方格中，任意五个球在一条直线或斜线上，就中奖了！
: 请问：中奖的概率有多大？
| a*********h
发帖数: 71 | 4 Let A_i, for 1<=i<=12, be the set of outcomes with the five squares on th i-
th line filled with 5 balls. We need to compute |A_1 U A_2 U ... U A_12|. We
notice that the intersection of any four or more above different sets is
always empty since there are no any outcomes with 12 balls appearing on four
or more lines. However, there are outcomes with 12 balls appearing on three
or less lines; for instance, 12 balls can appear on two columns and one row
.
For the outcomes with 12 balls appearing on three lines, there are 2*C(5,2)*
C(5,1) of them with either two rows and one column or two columns and one
row, there are 2*C(5,1)*C(4,1) of them with one row and one column and one
diagonal, and there are 8 of them with two diagonals and one row/column.
For the outcomes with at least two lines, there are 2*C(5,2)*C(15,2) of them
with two rows or two columns, there are C(5,1)*C(5,1)*C(16,3)of them
with one row and one column, there are 2*5*C(16,3) of them with one diagonal
and
one row/column, and there are C(16,3) of them with two diagonals. For the
outcomes with 12 balls appearing on at
least one line, there are C(12,1)*C(20,7) of them.
Using the inclusion and exclusion principle, we have |A_1 U A_2 U ... U A_12
|= 12*C(20,7)-(2*C(5,2)*C(15,2)+C(5,1)*C(5,1)*C(16,3)+2*5*C(16,3)+C(16,3))+(
2*C(5,2)*C(5,1)+2*C(5,1)*C(4,1)+8)=
908128.
So 908128/C(25,12)=908128/5200300=0.17463. | y*****w
发帖数: 1350 | 5 You forgot to take out the duplicates in the following cases:
1) a combination of one horizontal line and another horizontal line (n=10*C(
15, 2)=10*105)
2) a combination of one vertical line and another vertical line (n=10*C(15,
2)=10*105)
3) a combination of one horizontal line and one vertical line (n=25*C(15, 2)
=25*105)
4) a combination of one line (horizontal or vertical) and one diagonal (n=20
*C(15, 2)=20*105)
The calculation steps would be:
1) There are 12 cases where you can get a straight line of 5 balls.
2) In one such case, for the rest of the balls, there are C(20, 7)
possibilities.
3) Then get the "total" number of possibilities is 12*C(20, 7).
4) The nubmer in Step 3) did not take into account the duplicates. Therefore
, the correct number of possibilities is 12*C(20, 7) - (10+10+25+20)*105 =
12*77520 - 6825 = 923415 by deducting those duplicates.
5) The total number of possibilities for 12 balls in a 5X5 panel is C(25, 12
) = 5200300
6) Consequently, the probability is 923415 / 5200300 = 0.17757.
【在 h********y 的大作中提到】
: 我的理解是，这个方格子总共12条线，每条线有5个格子，如果有5个球在一条线上，那
: 么有12种可能，而另外7个球会在剩下的20个方格里的任何方格，会有（20，7）个组合
: 。12个球在25个格子上的组合是（25，12），那么中奖概率就是12*（20，7）/（25，
: 12）=0.17888，excel算的
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