請列式+解釋!!!
A bag contains one red ball, one blue ball and two yellow balls. Two balls are drawn from the bag. Find the probability that the balls drawn are
(a) both yellow,
(b) red and yellow
(c) of different colours
PS,,, 有冇F3 學生可以解答我問題AR? 請註明你是否F3 學生, 如果係... 可唔可以留MSN 聯絡AR?
F.3 probability 概率 - 3
2007-06-30 6:32 pm
回答 (2)
2007-06-30 6:41 pm
✔ 最佳答案
圖片參考：http://i182.photobucket.com/albums/x4/A_Hepburn_1990/pluto.jpg?t=1183171074
a. P(both yellow)
= 2/4 X 1/3
= 1/6
b. P(red and yellow)
= 1/4 X 2/3 + 2/4 X 1/3
= 1/3
c. P(the balls are of different colours)
= P(red and yellow) + P(red and blue) + P(yellow and blue)
= 1/3 + (1/4 X 1/3 + 1/4 X 1/3) + (2/4 X 1/3 + 1/4 X 2/3)
= 1/3 + 1/6 + 1/3
= 5/6
參考: Myself~~~
2007-07-01 6:34 am
(a)P(yellow)
=2/4 x 1/3
=1/6
(b)P(red and yellow)
=P(red)xP(yellow) + P(yellow)xP(red)
=1/4 x 2/3 + 2/4 x 1/3
=1/3
(c)P(different colours)
=P(red and blue) + P(red and yellow) + P(blue and yellow)
=P(red)xP(blue)+P(blue)xP(red) + P(red)xP(yellow)+P(yellow)xP(red)+ P(blue)xP(yellow)+P(yellow)xP(blue)
=(1/4 x 1/3 x 2) + 1/3 + (1/4 x 2/3 + 2/4 x 1/3)
=5/6
P.S.我唔係F.3學生呀~但我係應屆會考生,希望可以幫到你啦!!
=2/4 x 1/3
=1/6
(b)P(red and yellow)
=P(red)xP(yellow) + P(yellow)xP(red)
=1/4 x 2/3 + 2/4 x 1/3
=1/3
(c)P(different colours)
=P(red and blue) + P(red and yellow) + P(blue and yellow)
=P(red)xP(blue)+P(blue)xP(red) + P(red)xP(yellow)+P(yellow)xP(red)+ P(blue)xP(yellow)+P(yellow)xP(blue)
=(1/4 x 1/3 x 2) + 1/3 + (1/4 x 2/3 + 2/4 x 1/3)
=5/6
P.S.我唔係F.3學生呀~但我係應屆會考生,希望可以幫到你啦!!
參考: 自己
收錄日期: 2021-04-23 21:30:06
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070630000051KK01044