Probability

圖片參考：https://s.yimg.com/lo/api/res/1.2/ZQit8r6YM3UMER1bF6V.OQ--/YXBwaWQ9dHdhbnN3ZXJzO3E9ODU-/http://i.imgur.com/ZH9B85L.gif

1.
(a)
Denote
X = Flower taken from Vase A
Y = Flower taken from Vase B

X can be rose (R) or carnation (C).
Y can be rose (R) or carnation (C).

Table of all possible outcomes (X, Y) is

－－－－－－－－－－－－－－－－－－－－－－－－－－－
　　　｜　　　　Ｙ＝Ｒ　　　　　　　　　Ｙ＝Ｃ
－－－＋－－－－－－－－－－－－－－－－－－－－－－－
Ｘ＝Ｒ｜（Ｘ，Ｙ）＝（Ｒ，Ｒ）　（Ｘ，Ｙ）＝（Ｒ，Ｃ）
　　　｜
Ｘ＝Ｃ｜（Ｘ，Ｙ）＝（Ｃ，Ｒ）　（Ｘ，Ｙ）＝（Ｃ，Ｃ）
－－－－－－－－－－－－－－－－－－－－－－－－－－－

(b)
The probability that the flowers chosen are of the same type
= P( (X, Y) = (R, R) or (X, Y) = (C, C) )
= P( (X, Y) = (R, R) ) + P( (X, Y) = (C, C) )
= P( X = R and Y = R ) + P( X = C and Y = C )
= P( X = R ) × P( Y = R ) + P( X = C ) × P( Y = C )
= (4/5) × (2/4) + (1/5) × (2/4)
= 8/20 + 2/20
= 10/20
= 1/2

✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀

2.
The probability that he gets a pass in his homework
= P( he gets C or above )
= P( he gets A ) + P( he gets B ) + P( he gets C )
= 3/20 + 4/20 + 8/20
= (3 + 4 + 8)/20
= 15/20
= 3/4


2015-07-06 21:29:34 補充：
For continuous data, if you divide into groups of
(10 - 14), (15 - 19), (20 - 24), (25 - 29), etc.

Look at the group of (15 - 19).

Class interval is (15 - 19).
Class mark = 17
Lower class limit = 15
Upper class limit = 19
Lower class boundary = 14.5
Upper class boundary = 19.5

2015-07-06 21:56:54 補充：
Q1 題目要求列表指出所有可能性。

(X,Y) = (花瓶 A 取出的花, 花瓶 B 取出的花)

花可以是玫瑰 (R) 或 康乃馨 (C)。

因此，(X,Y) 有四個可能性：(R,R), (R,C), (C,R), (C,C)。

總情況數 = 5 × 4 = 20
(R,R) 情況數 = 4 × 2 = 8
(R,C) 情況數 = 4 × 2 = 8
(C,R) 情況數 = 1 × 2 = 2
(C,C) 情況數 = 1 × 2 = 2

2015-07-06 22:02:41 補充：
Read my supplementary explanation.

oh my god i still cannot quite get q 1
but thanks for your explanation for no.2 and my ectra question!!!

2015-07-12 19:58:08 補充：
sorry 知足常樂 :'(
I am busy these days , so maybe i will read your further explanation later