✔ 最佳答案
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
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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.