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

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18 Jul 2009, 13:03

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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: probability:A committee of 3 has to be formed [#permalink]

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18 Jul 2009, 14:10

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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....

Re: probability:A committee of 3 has to be formed [#permalink]

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19 Jul 2009, 18:27

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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%

Re: probability:A committee of 3 has to be formed [#permalink]

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23 Jul 2009, 03:22

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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?

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

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20 Mar 2012, 15:19

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

Look for any of the three variants: Tom-notMary-notMary, notMary-Tom-notMary, notMary-notMary-Tom. The probability of one such variant is \(\frac{1}{6} * \frac{4}{5} * \frac{3}{4} = \frac{1}{10}\) . The answer to the question is \(3*\frac{1}{10} = \frac{3}{10}\) .

Why can't this be 5C3/ 6C3 = 1/2? Where did I make mistake when I applied combination formula approach? I know that this leads to wrong answer.

Here is why 5C3 is wrong: 5C3 is the total number of ways to select 3 people from 5. If Tom is one of the 5 people, 5C3 does not count ONLY those ways in which Tom will be included. It counts the number of ways in which 3 people can be selected from 5 people (including Tom). This will include selections which do not include Tom, and so overstate the number of ways of always including Tom in the selection.

The correct solution: Tom will always be selected, so select the other 2 people from the remaining 4 people. Number of ways = 1 * 4C2 = 6 Total number of ways to select 3 people from 6 = 6C3 = 20 Therefore probability = 6/20 = 3/10
_________________

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

Look for any of the three variants: Tom-notMary-notMary, notMary-Tom-notMary, notMary-notMary-Tom. The probability of one such variant is \(\frac{1}{6} * \frac{4}{5} * \frac{3}{4} = \frac{1}{10}\) . The answer to the question is \(3*\frac{1}{10} = \frac{3}{10}\) .

Why can't this be 5C3/ 6C3 = 1/2? Where did I make mistake when I applied combination formula approach? I know that this leads to wrong answer.

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}.

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

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07 Jun 2012, 22:45

another way

T , M , A , B, C, D be the six members where T = Tom and M= Mary

First spot : let first person selected be tom so = 1/6 = Tom is selected from 6 people so M, A, B, C, D remain

Second spot : 4/5= mary cannot be selected so only 4 people can be selected from 5 ,after that only 4 people remain

Third spot : 3/4 = Again Mary cannot be selected ,

now it is not necessary that T will be picked first , we can have _ T _ or _ _ T or T _ _ , so we have to make order of the same elements irrelevant . The non mary candidates are same so we can have only 3 cases i.e [ Non mary , Non mary , T ] and [non mary , T , Non mary] and [ T , Non mary , Non mary ] .so 3!/ 2! , which equals 3

Just like arranging the letters of the word [ ALL ] = ALL ,LAL, LLA . only 3 cases possible , which is 3!/ 2! ( divide by the number of same elements)

there fore : 1/6 * 4/5 *3/4 * 3!/2! = 3/10

Its good to know all methods in case you need them, cheers !!

Re: probability:A committee of 3 has to be formed [#permalink]

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03 Jul 2012, 10:49

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]

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03 Jul 2012, 21:20

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.

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}.

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

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20 Oct 2012, 21:50

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

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

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30 Oct 2012, 13:59

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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)

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

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06 Jan 2013, 12:09

Mary is not to be selected, and Tom is to be selected.

Thus tom and mary are out of picture in selection (tom selected, marry rejected) Problem boils to selecting other 2 members out of remaining 4 which can be done in 4C2 ways.

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

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10 Jan 2014, 09:08

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?

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

So Tom Yes and Mary NO

Then we will have to pick 2 other that will not include Mary

So we have 4C2 = 6

Now total number of options without restrictions were 6C3

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

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17 Nov 2014, 03:36

Does anyone know how the reverse combination and reverse probability approach would work here? Also, I hear there are elimination strategies for probability PS questions. Does anyone see any, here? ATB

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) 1/10

b) 1/5

c) 3/10

d) 2/5

e) 1/2

Please refer to the solutions above.
_________________

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

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22 Jul 2015, 07:12

Bunuel wrote:

chetan2u 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) 1/10

b) 1/5

c) 3/10

d) 2/5

e) 1/2

Please refer to the solutions above.

Hi Bunuel !

Why does order matter in this question? I mean we're concerned with the 3 people who would form the committee. IMO - membership matters not the order in which members are chosen. Opinions?

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) 1/10

b) 1/5

c) 3/10

d) 2/5

e) 1/2

Please refer to the solutions above.

Hi Bunuel !

Why does order matter in this question? I mean we're concerned with the 3 people who would form the committee. IMO - membership matters not the order in which members are chosen. Opinions?

Regards, SR

Hi, order does not matter in this Q, that is why we have taken combinations and not permutations.. 4C2/6C3..
_________________

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