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# A committee of 3 has to be formed randomly from a group of 6 people.

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A committee of 3 has to be formed randomly from a group of 6 people. [#permalink]

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23 Jul 2010, 08:56
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76% (02:08) correct 24% (01:04) wrong based on 29 sessions

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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?

A. $$\frac{1}{10}$$
B. $$\frac{1}{5}$$
C. $$\frac{3}{10}$$
D. $$\frac{2}{5}$$
E. $$\frac{1}{2}$$

(C) 2008 GMAT Club - m23#30

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-committee-of-3-has-to-be-formed-randomly-from-a-group-of-81051.html
[Reveal] Spoiler: OA

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Re: A committee of 3 has to be formed randomly from a group of 6 people. [#permalink]

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23 Jul 2010, 11:38
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 has to be formed randomly from a group of 6 people. [#permalink]

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02 Aug 2010, 04:00
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
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Re: A committee of 3 has to be formed randomly from a group of 6 people. [#permalink]

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02 Aug 2010, 17:14
1
KUDOS
[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 has to be formed randomly from a group of 6 people. [#permalink]

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02 Aug 2010, 17:22
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 has to be formed randomly from a group of 6 people. [#permalink]

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26 Dec 2011, 23:46
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 has to be formed randomly from a group of 6 people. [#permalink]

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27 Dec 2011, 03:31
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
ANS is C
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Re: A committee of 3 has to be formed randomly from a group of 6 people. [#permalink]

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03 May 2016, 01:33
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Re: A committee of 3 has to be formed randomly from a group of 6 people. [#permalink]

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03 May 2016, 04:28
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?

A. $$\frac{1}{10}$$
B. $$\frac{1}{5}$$
C. $$\frac{3}{10}$$
D. $$\frac{2}{5}$$
E. $$\frac{1}{2}$$

(C) 2008 GMAT Club - m23#30

You can solve this problem with direct probability approach as well. We need the probability of T, NM, NM (Tom, not Mary, not Mary). $$P(T, NM, NM)=3*\frac{1}{6} * \frac{4}{5} * \frac{3}{4}=\frac{3}{10}$$, we are multiplying by 3 since {T, NM, NM} scenario can occur in 3 different ways: {T, NM, NM}, {NM, T, NM}, {NM, NM, T}.

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-committee-of-3-has-to-be-formed-randomly-from-a-group-of-81051.html
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Re: A committee of 3 has to be formed randomly from a group of 6 people.   [#permalink] 03 May 2016, 04:28
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