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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. \(\frac{1}{2}\)
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Re M2330
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16 Sep 2014, 01:19
Official Solution: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: M2330
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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: M2330
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22 Oct 2014, 23:55
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: M2330
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24 Oct 2014, 01:01
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: M2330
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01 Jan 2015, 20:59
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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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: M2330
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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: M2330
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13 Mar 2016, 02:16
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: M2330
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28 Jul 2016, 13:40
Hi Bunuel
Why are we not multiplying by 3! instead of 3  those two NM's (NM1, NM2) also can be chosen differently.. so there should be total of 6 ways to select them .
Please clarify.



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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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Re: M2330
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07 Sep 2016, 13:34
this is not a permutation question, where is Tom position in the group of three doesnt matter
easiest way total combination 6c3
tom will occupy one spot and remaining 02 spots in the group will be filled from 04 person (05 not including mary) 1*4c2
so probability 4c2/6c3



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Re: M2330
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20 Sep 2016, 06:54
I did it correctly and the difficulty level changed from hard to medium



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Re: M2330
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22 Feb 2017, 11:58
mvictor wrote: 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 This is incorrect. Consider the six people to be {A, B, C, D, Tom, Mary} There are 10 combinations which include Tom, of which 6 do not include Mary: {A,B,Tom}; {B,C,Tom}; {C,D,Tom}; {A,C,Tom}; {A,D,Tom}; {B,D,Tom}; {A, Tom, Mary}; {B, Tom, Mary}; {C, Tom, Mary}; {D, Tom, Mary}. Per the same logic above, there are 10 combinations which include Mary, which means there are 10 combinations that do not include Mary. Combos with Tom only: 6 Combos with Mary only: 6 Combos with Tom and Mary: 4 Combos without Tom and Mary: 4



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Re: M2330
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22 Feb 2017, 12:20
shasadou wrote: 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. This isn't correct, there are only four combinations when the two are chosen together. For your approach, you'd have to do: 1(Mary only)(Tom and Mary)(without Tom or Mary) = 1 (1c1*1c0*4c2)/6c3  (2c2*4c1)/6c3  (2c0*4c3)/6c3 = 6/20=3/10 Combos with Tom only: 6 Combos with Mary only: 6 Combos with Tom and Mary: 4 Combos without Tom and Mary: 4



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Re: M2330
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13 Jul 2017, 18:36
I've solved it this way: Probability that Tom will be chosen is to fill up one of the three seats is 3/6, and probability that Mary will not be chosen is 3/5 (5 is the remaining seats available). When I multiply those two probabilities I get the following: 3/6 x 3/5 = 3/10 Answer C



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Re: M2330
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27 Mar 2018, 06:00
+1 for option C. The required probability is : 4c2/6c3.
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Re: M2330
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29 Mar 2018, 16:23
Group of People 6 >>>> Committee needs to be formed of 3 Members. Mary is not required in the committee so keep her out. Group of people =61 (subtracting Mary from total) = 5. Now we have to keep Tom. Probability of Tom for sure \(\frac{1}{5}\). Mary is gone and now Tom has taken place in committee of 3 so, Group of people=51 (subtracting 1 for Tom as now he is committee) Now the remaining 2 post of the committee can been filed in \(\frac{2}{4}\) ways. Total probability = \(\frac{1}{5}\) * \(\frac{2}{4}\) = \(\frac{1}{10}\). Wrong. We need to multiple Total probability by 3 as Mr. Tom can be selected at any three places. _T_, T_ _ or _ _ T. Hence, total probability = 3*\(\frac{1}{10}\) = \(\frac{3}{10}\). (C).
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Re: M2330
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13 Dec 2018, 12:08
Bunuel wrote: Official Solution:
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 I am getting 47/120 as the probablity Following is my solution. Please let me know where am I going wrong 3 Ways to select the team 1. {T, NM,NM} > p1 = (1/6) * (4/5) * (3/4) 2. {NM,T,NM} > p2 = (5/6) * (1/5) * (3/4) 3. {NM, NM, T} > p3 = (5/6 * (4/5) * (3/4) Total P = p1+p2+p3 = 47/120



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Re: M2330
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13 Dec 2018, 21:42
Amitrath1 wrote: Bunuel wrote: Official Solution:
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 I am getting 47/120 as the probablity Following is my solution. Please let me know where am I going wrong 3 Ways to select the team 1. {T, NM,NM} > p1 = (1/6) * (4/5) * (3/4) 2. {NM,T,NM} > p2 = ( 5/6) * (1/5) * (3/4) 3. {NM, NM, T} > p3 = ( 5/6 * ( 4/5) * (3/4) Total P = p1+p2+p3 = 47/120 In the second and third cases when you have Not Mary, it should also mean not Tom because you have separate spot for Tom there. You'll get correct answer if you correct that.
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