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M23-30

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Bunuel
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M23-30 [#permalink] New post   16 Sep 2014, 01:19
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63% (01:33) correct 37% (01:49) wrong based on 143 sessions
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A committee of 3 people must be formed randomly from a group of 6 individuals. If Tom and Mary are both part of this group of 6, what is the probability that Tom will be chosen for the committee, while Mary will not?

A. \(\frac{1}{10}\)
B. \(\frac{1}{5}\)
C. \(\frac{3}{10}\)
D. \(\frac{2}{5}\)
E. \(\frac{1}{2}\)
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C
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minhaz3333
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Re: M23-30 [#permalink] New post   15 Aug 2016, 02:44
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In my way,

I found this to be the easiest solution for the question.
ways of choosing 3 people out of 6 = 6C3
ways of choosing only Tom from Tom and Mary = 1
and choosing remaining 2 from remaining 4 (excluding tom and Mary) = 4C2

Probability of choosing 3 member including Tom but not Mary =
1 x 4C2
--------- = 3/10
6C3
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Bunuel
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Re: M23-30 [#permalink] New post   22 Oct 2014, 23:55
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manish2014 wrote:
I am getting ans as 1/2

we have to find the probability of getting (T,NM,NM)

NM,NM - 5C2=10

and total ways- 6C3=20

therefore - 10/20 =1/2


It should be 4C2: choosing 2 out out of 4 (6 - Tom - Mary).
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Re: M23-30 [#permalink] New post   24 Oct 2014, 01:01
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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. \(\frac{1}{2}\)


Committee of 3 can be formed out of 6 people in 6C3 ways or 20 ways

Now Consider committee in which Tom is already a member so out of remaining 4 members (excluding Mary)..2 can be selected in 4C2 ways = 6 ways..

So probability 6/20 or 3/10
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Re M23-30 [#permalink] New post   16 Sep 2014, 01:19
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Official Solution:

A committee of 3 people must be formed randomly from a group of 6 individuals. If Tom and Mary are both part of this group of 6, what is the probability that Tom will be chosen for the committee, while 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 having Tom (T), not Mary (NM), and another person who is not Mary (NM) on the committee (T, NM, NM).

The probability of this combination (T, NM, NM) can be calculated as follows:

\(P(T, NM, NM)=3*\frac{1}{6} * \frac{4}{5} * \frac{3}{4}=\frac{3}{10}\)

We multiply by 3 because the (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: M23-30 [#permalink] New post   22 Oct 2014, 21:19
I am getting ans as 1/2

we have to find the probability of getting (T,NM,NM)

NM,NM - 5C2=10

and total ways- 6C3=20

therefore - 10/20 =1/2
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Re: M23-30 [#permalink] New post   01 Jan 2015, 20:59
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i solved it this way:
6C3 = 20
combinations to choose Tom = 3C1 = 3
combinations not Mary = 3C1 = 3

we add these two 3+3 = 6
6/20 = 3/10
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Re: M23-30 [#permalink] New post   16 Jan 2016, 02:03
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6!/3!3! = 20 total possible combinations. 5!/3!2! = 10 combinations when the 2 are chosen together. 4!/3!1! = 4 comb when the 2 are out.

hence 1 - 14/20 = 3/10 combinations when either is in the team.
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Re: M23-30 [#permalink] New post   13 Mar 2016, 01:10
The solution makes sense, but can someone please let me know what I did wrong?

Total number of ways = 6C3 = 6*5*4/(3*2) = 20 ways -- this represents order doesn't matter

Total number of ways to select Tom, Not Marie, Not Marie = 1*4*3/(3*2) = 2 ways -- divide by 3*2 b/c order doesn't matter (to match denominator)

Thus 2/20 = 1/10; why is this wrong??
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Re: M23-30 [#permalink] New post   13 Mar 2016, 02:16
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happyface101 wrote:
The solution makes sense, but can someone please let me know what I did wrong?

Total number of ways = 6C3 = 6*5*4/(3*2) = 20 ways -- this represents order doesn't matter

Total number of ways to select Tom, Not Marie, Not Marie = 1*4*3/(3*2) = 2 ways -- divide by 3*2 b/c order doesn't matter (to match denominator)

Thus 2/20 = 1/10; why is this wrong??


Hi,

you have gone wrong here..


in initial 6C3, 3*2 was there because 3 people were selected and these 3 can be arranged in 3! ways..
But in second case, you have already selected 1 out of 3, and you have to select 2 out of remaining 4..
these two can be arranged in 2! so you have to divide by 2! and not 3!=3*2..

second case is 6 people- 1 already selected and 1 left out ..


2 to be picked up from 4..
so 4C2=4!/2!2!=6..
ans 6/20=3/10
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Re: M23-30 [#permalink] New post   18 Apr 2023, 15:03
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M23-30 [#permalink]
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