URGENT@@HELP!! 20分

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.)

SOR 是1-119/429 =310/429

2015-01-20 01:36:56 補充：
我不肯定!!!!!!