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A committee of 3 has to be formed randomly from a group of 6 [#permalink]
 18 Jul 2009, 14:03
Question Stats:
62% (02:21) correct
37% (01:19) wrong based on 56 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? 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
Last edited by Bunuel on 04 Jul 2012, 02:11, edited 1 time in total.
Edited the question, added the answer choices and OA
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Re: probability:A committee of 3 has to be formed [#permalink]
18 Jul 2009, 15:10
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3C6 = 20, which means that 20 groups of 3 are possible to be formed. T _ _ => 2C5 = 10, which means that 10 groups of 3 will have TOM Now here is the tip. When a problem ask you to calculate the number of times that something NOT happen, you calculate the number of times that something HAPPENS, and then calculate the difference: T M _ => = 4, which means that 4 groups will have TOM and MARY. Thus 10 - 4 = 6, which means that only 6 groups will have TOM but NOT mary The probability is 6/20 = 30% Is this the right answer? You know, everybody is human... Maybe I did some mistake in the process.... 
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Re: probability:A committee of 3 has to be formed [#permalink]
19 Jul 2009, 19:27
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Number of ways of chosing 3 out of 6 people = 6C3 = 20.
Now T should always be there in the committe. So we have to chose 2 people out of 5, but we are given that Mary should NOT be in the committe. So all we need to do is select 2 people out of 4 people = 4C2 = 6.
Hence probability of Tom in the committte and Mary NOT in the committe = 6/20 = 3/10 = 30%
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Re: probability:A committee of 3 has to be formed [#permalink]
20 Jul 2009, 16:18
Well, this last one was faster.
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Re: probability:A committee of 3 has to be formed [#permalink]
21 Jul 2009, 21:12
Followed the same technique as sdrandom1, quick solution.
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Re: probability:A committee of 3 has to be formed [#permalink]
23 Jul 2009, 04:22
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Sample space for selecing committe of 3 people out of 6 6C3 Number of way, both TOM AND mary are excluded and committe can be formed 4C3 Number of way, when both are there in teh committe 4C1, So probability when either of them are there are not= (6C4-(4C3+4C1))/6C3 =60% In case, only one has to be there, 60%/2 = 30% pmal04 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?
(C) 2008 GMAT Club - m23#30
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Re: probability:A committee of 3 has to be formed [#permalink]
03 Jul 2012, 11:49
can somebody tell me where am I lost..?? when I choose a group of 3 out of 6 it should be 6c3 / 2 as when you are selecting 3 people out from a group the other 3 are automatically a group. another area of concern, I agree that T has to be in a group always than that leaves me to option of selecting 2 from 4c2 but it is actually 6c1 * 4c2.. what am I missing here..can someone please help.?
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Re: probability:A committee of 3 has to be formed [#permalink]
03 Jul 2012, 22:20
Quote: when I choose a group of 3 out of 6 it should be 6c3 / 2 as when you are selecting 3 people out from a group the other 3 are automatically a group We are choosing one group....of 3 people.....while choosing 3 people...if other are left out....it doesnt matter.....its something like....if teacher finds 5 people chit chatting at back bench and ask them to leave the class...the 5 student standing outside doesnt make another class...they are just left out. Quote: I agree that T has to be in a group always than that leaves me to option of selecting 2 from 4c2 but it is actually 6c1 * 4c2 I am not sure how you came to 6c1. In our group, Tom is already there.So we are left with 5 people. Mary cannot be in the choosen people....so we are left with 4 people...from whom we need to select the two people..so 4C2 = 6 ways. Hope this helps.
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Re: A committee of 3 has to be formed randomly from a group of 6 [#permalink]
04 Jul 2012, 02:11
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} We need to find 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}, or {NM, NM, T}. Answer: C.
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Re: A committee of 3 has to be formed randomly from a group of 6 [#permalink]
20 Oct 2012, 22:50
Bunuel 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}
We need to find 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}, or {NM, NM, T}.
Answer: C. Elaborating on the above: We can look at this problem as a draw of 3 committee members without replacement. The draws can be done in the following possible 3 ways with only Tim in it and not Mary in it ( these are our favorable outcomes of the experiment) - {T, NM, NM}, {NM, T, NM}, or {NM, NM, T} Sum of the probabilities of the favorable outcomes is as calculated by Bunuel above P{T, NM, NM} = 1/6 * 4/5 * 3/4 P{NM, T, NM} = 4/6 * 1/5 * 3/4 ( please understand that NM in first draw also means not Tim) P{NM, NM, T} = 4/6 * 3/5 * 1/4
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Re: A committee of 3 has to be formed randomly from a group of 6 [#permalink]
30 Oct 2012, 14:59
For anyone who's still confused, here's some more clarification: For the subcommittee, you basically want to choose Michael, then choose 2 people who do not include Mary. Order matters - meaning Michael in the 1st slot should be counted separately from Michael in the 2nd and 3rd slots. (Out of 1 available Michael, pick that 1 Michael) * (Out of remaining 4 people excluding Mary, pick 2 people for the 2nd and 3rd slots)
= -----------------------------------------------------------------------------------------------------------------------------------------------------------------
(Out of 6 available people pick 3 for the committee)= (1C1) * (4C2) / (6C3) = (1) * (4*3 / 2) / (6*5*4 / 3!) = 6 / 20 = 3 / 10 = 30%
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Re: A committee of 3 has to be formed randomly from a group of 6
[#permalink]
30 Oct 2012, 14:59
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