# URGENT@@HELP!! 20分

### 2 個解答

• Tony
Lv 4
6 年前
最愛解答

1.(a)

the probability that the committee consists of no juniors

= 9C5 / 15C5

(ps. 9 freshmen and sophomores, 15 students, 5 picked students)

= 6/143

= 0.0420 (3 sig. fig.)

the probability that the committee consists of no freshmen

= 11C5 / 15C5

= 2/13

= 0.154 (3 sig. fig.)

1(b) the probability that the committee consists of no juniors and no freshmen

= 5C5 / 15C5

= 1/3003

= 0.000333 (3 sig. fig.)

1(c)

Before I teach you how to do this type of question,

I will talk about the wrong concept that students always have

wrong concept

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the number of ways of having

at least one member from each year group in the committee

= 4C1 * 5C1 * 6C1 * 12C2 = 7920

the student with this wrong concept commit double counting!!!!

1: A (freshman), B (sophomore), C (junior), D (freshman), E (freshman)

2: D (freshman), B (sophomore), C (junior), A (freshman), E (freshman)

1 and 2 are the same as 1 and 2 are formed by the same people(A,B,C,D,E)

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correct concpet!!!

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The ways of forming a committee:

(1): all members are from the SAME year group

(2): all members are from the EXACTLY TWO DIFFERENT year group

(3): all members are from the (EXACTLY THREE) DIFFERENT year group

so, the number of ways of having

at least one member from each year group in the committee

= all members - (1) - (2)

= 15C5 - (0 + 1 + 6C5) - (11C5 + 10C5 + 9C5)

(ps. 11C5: no freshmen, 10C5: no sophomores, 9C5: no juniors)

= 2156

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the probability that in the committee there is at least one member

from each year group

= 2156/3003

= 0.718 (3 sig. fig.)

資料來源： have taken a probability course in university
• 6 年前

SOR 是1-119/429 =310/429

2015-01-20 01:36:56 補充：

我不肯定!!!!!!