Probability

2015-07-07 5:12 am
1. There are some flowers in vases A and B . In vase A, there are 4 roses and 1 carnation. in vase b , there are 2 roses and 2 carnations. A flower is chosen from each vase at random.
(a) list all possible outcomes in a table.
(b) find the probability that the flowers chosen are of the same type .

2. The following table shows the grades of 20 students in a Visual Art homework.
Grade
A B C D E
no. of students
3,4,8,3,2

The passing grade of the homework is C. If a student is chosen at random from these 20 students, find the probability that he gets a pass in his homework.
更新1:

sorry therre is sth that flashed my mind wt's the diff. btwn class interval and class boundary? I forgot .......

更新2:

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

回答 (2)

2015-07-07 5:26 am
✔ 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

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

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.
2015-07-07 6:01 am
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


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