# 查看完整版本 : M1概率題求解答

[attach]13861756[/attach]

（996/1000) * (2/999) * (2/998)?

[[i] 本帖最後由 寒冬委屈的白襪 於 2022-12-24 18:11 編輯 [/i]]

top11 2022-12-24 22:40

[quote]原帖由 [i]好望角234[/i] 於 2022-12-24 17:53 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=554724801&ptid=30919164][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]
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To draw 3 numbers from the set {1, 2, 3, ..., 999, 1000}, if the 3 numbers can form an arithmetic sequence, the common difference (d) can be as small as 1 and as large as 499.

For d = 1, the cases are {1,2,3}, {2,3,4}, {3,4,5}, ... , {998,999,1000}.
There are 998 cases.

For d = 2, the cases are {1,3,5}, {2,4,6}, {3,5,7}, ..., {996,998,1000}.
There are 996 cases.

For d = 3, the cases are {1,4,7}, {2,5,8}, {3,6,9}, ..., {994,997,1000}.
There are 994 cases.

...

For d = 499, the only cases are {1,500,999} and {2,501,1000}.
There are 2 cases.

Therefore, in total the number of cases (to form arithmetic sequences) is
998 + 996 + 994 + ... + 2
= (998 + 2) × 499 ÷ 2
= 249500.

The total number of ways to draw 3 numbers from 1000 numbers is
C(1000, 3)
= 1000 × 999 × 998 / 3!
= 166167000

The required probability is
249500/166167000
= 1/666
= 0.001501502