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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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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. \(\frac{1}{2}\)

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?

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.

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

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This post received 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.

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. _________________

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 so answer is 4C2/6C3=6/20=3/10 ANS is C

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

Expert's post

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}.

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