Probability question

2007-12-27 5:47 am
There are four pairs of black socks and four pairs of white socks in a drawer. A sock is drawn from the drawer at random each time without replacement.

(a)If two socks are drawn, what is the probability that they are in same color?

(b)At least how many socks should be drawn to ensure that there is a pair of socks in the same color?
更新1:

show the step

更新2:

answer : (a) P = 7/15 (b) 3 the 3th answer is right

回答 (3)

2007-12-27 6:19 am
✔ 最佳答案
(a) P(socks are in same color) = 8/16 x 7/15 + 8/16 x 7/15 = 7/15

(b) 3 socks

解釋:
(a) 首先抽一隻襪,跟住再抽一隻。(抽第二隻前唔會放番第一隻襪落櫃桶。)
假設第一隻係白色襪: 8/16機會抽到白色。
第二隻一定要抽白色襪: (8-1)/(16-1) = 7/15。
第一隻要抽白襪,第二隻又要抽白襪係同時發生。
於是8/16乘7/15。 ←抽到一對白色襪的probability.

相反,如果第一隻抽到黑色襪: P = 8/16。
第二隻要抽到黑色襪: 7/15。
抽到一對黑色襪的probability: 8/16 x 7/15。

所以抽到同色襪的機會
= 抽到一對白色襪的probability + 抽到一對黑色襪的probability
= 8/16 x 7/15 + 8/16 x 7/15
= 7/15

(b)最少抽多少隻襪先確保一定有雙同色襪?
現在有兩隻色的襪,所以每次只有兩個可能:抽到黑色襪or白色襪。
抽三隻襪的所有組合:
1.白白白
2.白白黑
3.白黑白
4.白黑黑
5.黑白白
6.黑白黑
7.黑黑白
8.黑黑黑
我們可觀察到以上任何一組都有一對同色襪。
當抽三隻時,至少兩隻襪是同色。
因此答案是3 socks。
2007-12-27 6:32 am
a)

P (same color of socks)

= 4/8 x 3/7 x 2 .....分母8即會8隻socks,唔可以用一對來計,因為現在是一隻隻抽,唔係一對對,所以分母唔可以係4,x2係因為第一隻係黑,第二隻一定都要係黑(白既都一樣),而抽左一隻,個drawer就只得7隻socks,所以要係分子分母同時-1 , 可以唔寫x2 ,改為4/8 x 3/7 + 4/8 x 3/7 都可以 , 「+」有「或」的意思

= 3/7



b)

P (at least socks in the same color)

= 4/8 x 3/7 x 2/6 x 1/5 x2 ....第一隻抽黑,第二隻一定要抽到黑,第三隻都係,如此類推...抽到冇晒黑的為止 ,抽完就抽白色 ,同抽黑既次序係一樣的 ,要注意的是 ,抽第二隻時要同時將分子分母-1

= 1/70 x 2

= 1/35


2007-12-26 22:39:07 補充:
At least how many socks should be drawn to ensure that there is a pair of socks in the same color? ans.: 6 pairsBB , BB , BB / WW ,WW ,WW BW , BW , BW ,BW / BW , BW ,BW ,BW 回答上面位人兄 ,你解錯題吧 ,係「一隻抽完再一隻咁抽 」,冇話話抽三隻!
2007-12-27 5:51 am
If two socks are drawn, what is the probability that they are in same color?
A: 1+1=2


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