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cwk1668 2017-2-20 12:10 PM

概率問題

hkpal 2017-2-20 02:45 PM

cwk1668 2017-2-20 03:16 PM

(a)唔識
(b)15C4x11C4x7C4/(15C4)^3  唔知係唔係

[3] 2017-2-20 03:37 PM

[quote]原帖由 [i]cwk1668[/i] 於 2017-2-20 03:16 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=456716459&ptid=26459703][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]
(a)唔識
(b)15C4x11C4x7C4/(15C4)^3  唔知係唔係 [/quote]
[img]http://sciencesoft.at/lpng/af59ed0029f4fe0441e6c899159065055257c11.png&size=100[/img]

[3] 2017-2-20 03:46 PM

b right, 其實a差不多

cwk1668 2017-2-20 03:50 PM

[quote]原帖由 [i][3][/i] 於 2017-2-20 03:37 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=456717392&ptid=26459703][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]

[img]http://sciencesoft.at/lpng/af59ed0029f4fe0441e6c899159065055257c11.png[/img]&size=100 [/quote]

[3] 2017-2-20 04:01 PM

[quote]原帖由 [i]cwk1668[/i] 於 2017-2-20 03:50 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=456717985&ptid=26459703][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]

combinations for choice of 4=15C4
total number of combinations for 3 people=(15C4)^3
p(same choice for all 3)=15C4/(15C4)^3=1/(15C2)^2

cwk1668 2017-2-20 04:08 PM

[quote]原帖由 [i][3][/i] 於 2017-2-20 04:01 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=456718678&ptid=26459703][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]

combinations for choice of 4=15C4
total number of combinations for 3 people=(15C4)^3
p(same choice for all 3)=15C4/(15C4)^3=1/(15C2)^2 [/quote]

[3] 2017-2-20 04:13 PM

b right

chdse 2017-2-20 05:54 PM

Thx