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A committee of 3

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Director
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A committee of 3 [#permalink] New post 23 Jul 2010, 07:56
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Question Stats:

50% (02:31) correct 50% (01:20) wrong based on 6 sessions
A committee of 3 has to be formed randomly from a group of 6 people. If Tom and Mary are in this group of 6, what is the probability that Tom will be selected into the committee but Mary will not?



.1

.2

3/10

.4

.5
[Reveal] Spoiler: OA

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Re: A committee of 3 [#permalink] New post 23 Jul 2010, 10:38
rxs0005 wrote:
A committee of 3 has to be formed randomly from a group of 6 people. If Tom and Mary are in this group of 6, what is the probability that Tom will be selected into the committee but Mary will not?



.1

.2

3/10

.4

.5


I think this one is already solved here: probability-a-committee-of-3-has-to-be-formed-81051.html

But anyway, my try:

Assuming that Tom is selected and Ann must not be selected, you get to choose: 4C2 = 6 possibilities, over the total number of possibilities which are 6C3 = 20.

That gives us 6/20 --> 3/10.
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Re: A committee of 3 [#permalink] New post 02 Aug 2010, 03:00
two ways to solve this kind of problem

way 1. count the number of event needed and number of event available

number of events available is 6*5*4/3*2*1 (set without order-Princeton review). Number is 20

number of events needed is: we take Tom and 2 persons from 4 person. there are 4*3/2 (set without order). number is 6

result is 6/20

way 2: separate into cases, find P of each case and multiple or plus together.

case 1: find P (A and first person and second person)

P1=(1/6)*(4/5)*(3/4)=1/10

case 2, find P (first person and A and second person

P2=(4/6)*(1/5)*(3/4)=1/10

case 3, find P (first person and second person and A)

P3=(4/6)*(3/5)*(1/4)=1/10

P=p1+p2+p3=3/10
Senior Manager
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Re: A committee of 3 [#permalink] New post 02 Aug 2010, 16:14
[quote="rxs0005"]A committee of 3 has to be formed randomly from a group of 6 people. If Tom and Mary are in this group of 6, what is the probability that Tom will be selected into the committee but Mary will not?



.1

.2

3/10

.4

.5

That's a good one.

C (3,6) - all the possible arrangements

C(2,4) is the number of ways the committe with Tom and without Mary.

P = C(2,4)/(C 3,6) = 3/10
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Re: A committee of 3 [#permalink] New post 02 Aug 2010, 16:22
thangvietnam wrote:
two ways to solve this kind of problem

way 1. count the number of event needed and number of event available

number of events available is 6*5*4/3*2*1 (set without order-Princeton review). Number is 20

number of events needed is: we take Tom and 2 persons from 4 person. there are 4*3/2 (set without order). number is 6

result is 6/20

way 2: separate into cases, find P of each case and multiple or plus together.

case 1: find P (A and first person and second person)

P1=(1/6)*(4/5)*(3/4)=1/10

case 2, find P (first person and A and second person

P2=(4/6)*(1/5)*(3/4)=1/10

case 3, find P (first person and second person and A)

P3=(4/6)*(3/5)*(1/4)=1/10

P=p1+p2+p3=3/10



As you noticed that all 3 probabilities are the same and they are the same for a reason.
Consider this example what is the probablity of selecting the integer 1 out of 4 different numbers, it's 1/4 and what is the probability of selecting 2 out of 4 different numbers, it's also 1/4.

The same principle is applicable to this problem. The order in which you select a person, first, second or third does not matter, the probability is the same.
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Re: A committee of 3 [#permalink] New post 26 Dec 2011, 22:46
How did you guys come up with 4C2...? I do not understand this step...please enlighten me:)
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Re: A committee of 3 [#permalink] New post 27 Dec 2011, 02:31
we have 6 people
Tom is always selected so we are left with 2 people to choose from rest 5 people
since marry is never selected so question left is selecting 2 people from rest 4 people
so answer is 4C2/6C3=6/20=3/10
ANS is C
Re: A committee of 3   [#permalink] 27 Dec 2011, 02:31
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