Probability question

由 smb » 2010年12月12日 10:52:57 (p#2020694)

A purse contains two $0.5, three $1 and two $2 coins. If two coins are taken at random from the purse, find the probability that the total value of the coins is greater than $2.6

由 Snoopy@J&H » 2010年12月12日 11:22:42 (p#2020711)

唔知正唔正確
(1/3x1/2)x2=1/3

由園田海未 » 2010年12月12日 11:43:06 (p#2020722)

佢問七個銀抽兩個出黎多過$2.6 而最大面額個銀只係$2 其實個$0.5喺呢個case只係多餘
P (Greater than $2.6)

calculations：隱藏內容，按一下這裏顯示: = P ($2 at first) * P ($2 at second) + P ($2 at first) * P ($1 at second)
= 2/7 * 1/6 + 2/7 * 3/6
= 2/42 + 6/42
= 4/21

2樓條式點做出黎?

Snoopy@J&H :唔知正唔正確
(1/3x1/2)x2=1/3

由 smb » 2010年12月12日 13:35:27 (p#2020771)

和菓子 :佢問七個銀抽兩個出黎多過$2.6 而最大面額個銀只係$2 其實個$0.5喺呢個case只係多餘
P (Greater than $2.6)
calculations：隱藏內容，按一下這裏顯示

= P ($2 at first) * P ($2 at second) + P ($2 at first) * P ($1 at second)
= 2/7 * 1/6 + 2/7 * 3/6
= 2/42 + 6/42
= 4/21

2樓條式點做出黎?
Snoopy@J&H :唔知正唔正確
(1/3x1/2)x2=1/3

答案係1/3

但snoopy兄條式點出黎架?

由 GK9141@93A » 2010年12月12日 16:43:27 (p#2020821)

我就有另外兩個方法去計...

1)Combustion-中六Applied Maths
[($1&$2所有組合)+($2&$2所有組合)]/(所有組合,即包括$0.5,$1,$2)

[(2C1)(3C1)+(2C2)(3C0)]/(7C2)
=1/3

2)Probabitity-中三Maths
(第一次抽$2&第二次抽$1或$2既prob)+(第一次抽$1&第二次抽$2既prob)

2/7x4/6+3/7x2/6
=1/3

恕小弟不才,真係唔明點解2(1/2x1/3)可以出到答案 [banghead]

由 GX7271 » 2010年12月12日 19:43:23 (p#2020940)

smb :A purse contains two $0.5, three $1 and two $2 coins. If two coins are taken at random from the purse, find the probability that the total value of the coins is greater than $2.6

3/7 x 2/6（先抽 $1, 再抽 $2）+ 2/7 x 3/6（先抽 $2, 再抽 $1）+ 2/7 + 1/6（兩次都抽 $2）= 1/3