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From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
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01 May 2016, 11:24
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From a Group of 8 People, Including George and Nina, 3 people are to be selected at random to work on a certain project. What is the probability that 3 people selected will include George but not Nina A 5/56 B 9/56 C 15/56 D 21/56 E 25/56
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Re: From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
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01 May 2016, 12:28
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Number of ways of selecting 3 people out of 8 people = 8C3
In the three members George will always be there in the team. At this step we have vacancy for 2 more members and 7 members are available. Lina cannot be there in the team. So 2 members have to be selected and the number of available members = 7  Lina = 6
Number of ways to form a 3 member team that includes George and excludes Lina = 6C2
Probability = 6C2/8C3 = 15/56
Answer: C



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Re: From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
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02 May 2016, 02:12
mohitmohit11 wrote: From a Group of 8 People, Including George and Nina, 3 people are to be selected at random to work on a certain project. What is the probability that 3 people selected will include George but not Nina
A 5/56 B 9/56 C 15/56 D 21/56 E 25/56 Overall group = 8 people including George and Nina Restrictions: Nina is not included and George is included. This means we have on position (George) filled already and we need to select 2 people from the remaining 7 people. But Nina is not to be included, hence we need to select 2 people from the remaining 6 people. Number of ways: 6C2 = 15 Total number of ways of selecting 3 people out of 8 = 8C3 = 56 Probability of selecting the desired group = 15/56 Correct Option: C



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Re: From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
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02 May 2016, 03:34



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From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
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Updated on: 19 Nov 2016, 08:19
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G _ _  N _ _ _ _ _
This is the configuration that we want. Let's choose them one at a time.
In the first group, we want George and then 2 other people who are not Nina. That's \(\frac{1}{8}*\frac{6}{7}*\frac{5}{6}\), with 3 ways to arrange George in that group. Notice that the order/selection of the second group doesn't matter at all since we've already accounted for them in the first group.
In total, that's \(\frac{3*6*5}{8*7*6}\) = \(\frac{3*5}{8*7}= \frac{15}{56}\) .
Answer:C
Originally posted by eaze on 05 May 2016, 10:24.
Last edited by eaze on 19 Nov 2016, 08:19, edited 1 time in total.



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Re: From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
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05 May 2016, 10:42
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Let's consider A,B,C,D,E,F,G,H
Proba = \(\frac{Number of outcomes with A and not B }{Total number of outcomes}\)
Total number of outcomes = 3 from 8 = \(\frac{8 * 7 * 6 }{3 * 2 * 1}\) = 56
Number of outcomes with A and not B
There is 3 cases :
A_ _ : \(\frac{1 * 6 * 5 }{3 * 2 *1}\) = 5
_A_ : \(\frac{6 * 1 * 5 }{3 * 2 *1}\) = 5
_ _A : \(\frac{6 * 5 * 1 }{3 * 2 *1}\) = 5
So Proba = \(\frac{Number of outcomes with A and not B }{Total number of outcomes}\)
= \(\frac{15 }{56}\)



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From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
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18 Jun 2016, 09:48
Bunuel, Karishima, someone please!
I don't understand the Combinations method, it makes no sense to me what so ever.
So I am trying to use Venn diagram here.
So: P(N)+P(G)P(N and G) = 1
GGeorge Nnina We want to know George only, so no Nina and no overlap George plus Nina.
P(N)=1P(not N) \(1 \frac{7}{8}*\frac{6}{7}*\frac{5}{6} = \frac{3}{8}\) P(G)=1P(not G) \(1 \frac{6}{7}*\frac{5}{6}*\frac{4}{5} = \frac{3}{7}\) P(N and G)= P(A) * P(B) \(\frac{9}{56}\)
So: 1P(N)+P(N and G) = P(G)
\(1\frac{56}{56}\frac{21}{56}+\frac{9}{56}\) SHOULD BE P(G), but it's \(\frac{26}{56}\)
How come????? What is wrong here?...
Thanks!



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Re: From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
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18 Jun 2016, 12:58
iliavko wrote: Bunuel, Karishima, someone please!
I don't understand the Combinations method, it makes no sense to me what so ever.
So I am trying to use Venn diagram here.
So: P(N)+P(G)P(N and G) = 1
GGeorge Nnina We want to know George only, so no Nina and no overlap George plus Nina.
P(N)=1P(not N) \(1 \frac{7}{8}*\frac{6}{7}*\frac{5}{6} = \frac{3}{8}\) P(G)=1P(not G) \(1 \frac{6}{7}*\frac{5}{6}*\frac{4}{5} = \frac{3}{7}\) P(N and G)= P(A) * P(B) \(\frac{9}{56}\)
So: 1P(N)+P(N and G) = P(G)
\(1\frac{56}{56}\frac{21}{56}+\frac{9}{56}\) SHOULD BE P(G), but it's \(\frac{26}{56}\)
How come????? What is wrong here?...
Thanks! Hi iliavko, Why are you using Venn diagram ? This tool is rather used for sets questions. OK ? The question asks the probability of choosing a group of 3 people from 8, including A and excluding B. So, the basic proba formula is P = (Number of outcomes where the event occurs) / (Number total of outcomes) Number of outcomes where the event occursStage 1 : Number oy ways to choose A = 1 Stage 2 : Number of ways to choose 2 people form 6 (8, minus A, minus B since you want A and no B) = 15 Number of outcomes where the event occurs = 15 Number total of outcomesNumber of ways choose 3 people from 8 = 56 So P = 15/56 OK ?



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Re: From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
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18 Jun 2016, 16:47
Hi Alex,
Yes but Venn diagram can be used here as far as I understand... What you explained is the combinations method and yes its simple and people use it all the time, but I want to know why my logic with Venn diagram here isn't working.



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Re: From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
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18 Jun 2016, 20:17
iliavko wrote: Bunuel, Karishima, someone please!
I don't understand the Combinations method, it makes no sense to me what so ever.
So I am trying to use Venn diagram here.
So: P(N)+P(G)P(N and G) = 1
GGeorge Nnina We want to know George only, so no Nina and no overlap George plus Nina.
P(N)=1P(not N) \(1 \frac{7}{8}*\frac{6}{7}*\frac{5}{6} = \frac{3}{8}\) P(G)=1P(not G) \(1 \frac{6}{7}*\frac{5}{6}*\frac{4}{5} = \frac{3}{7}\) P(N and G)= P(A) * P(B) \(\frac{9}{56}\)
So: 1P(N)+P(N and G) = P(G)
\(1\frac{56}{56}\frac{21}{56}+\frac{9}{56}\) SHOULD BE P(G), but it's \(\frac{26}{56}\)
How come????? What is wrong here?...
Thanks! Hi, You are going wrong in execution of your DATA.. P(N)=1P(not N) \(1 \frac{7}{8}*\frac{6}{7}*\frac{5}{6} = \frac{3}{8}\) P(G)=1P(not G) \(1 \frac{6}{7}*\frac{5}{6}*\frac{4}{5} = \frac{3}{7}\) P(N and G)= P(A) * P(B) \(\frac{9}{56}\) we are looking for  What is the probability that 3 people selected will include George but not Nina?so It is P( G but not N) = P(G)  P(N and G), where P(G) is when George is there and P(N and G) is when both are together.. so Using your calculations = \(\frac{3}{7}  \frac{9}{56} = \frac{249}{56} = \frac{15}{56}\) Ofcourse it is better to do with combinations but you have used Venn correctly upto a point and then faultered in final execution.. Hope it helps you and this is what you were looking for..
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Re: From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
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19 Jun 2016, 02:35
Hi Chetan!
Cool!, Thank you so much for the clarification!
So I guess the way I did it I was doublecounting something somewhere?.. Anyways, what you said makes sense, just grab the P(G) subtract the overlap p(G and N) and get the "pure" P(G).
Thank you for your help!



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Re: From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
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21 Jun 2016, 12:22
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iliavko wrote: Hi Chetan!
Cool!, Thank you so much for the clarification!
So I guess the way I did it I was doublecounting something somewhere?.. Anyways, what you said makes sense, just grab the P(G) subtract the overlap p(G and N) and get the "pure" P(G).
Thank you for your help! Hi iliavko thanks for pointing out this point of view ... !! I was obviously missing something ! According to Vienn Diagramm : P(A and not B) = P(A)  P(A and B) = P(A)  P(A)*P(B) = P(A) * [1 P(B)] = P(A) * P(not B) P(A) = \(\frac{3}{8}\) P(not B) = \(\frac{5}{8}\) So, P(A) * P(not B) = \(\frac{5}{8}\) * \(\frac{3}{8}\) = \(\frac{15}{64}\) Can you explain why it's wrong ? Thank you !!



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Re: From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
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21 Jun 2016, 12:26
chetan2u wrote:
P(N)=1P(not N) \(1 \frac{7}{8}*\frac{6}{7}*\frac{5}{6} = \frac{3}{8}\) P(G)=1P(not G) \(1 \frac{6}{7}*\frac{5}{6}*\frac{4}{5} = \frac{3}{7}\) P(N and G)= P(A) * P(B) \(\frac{9}{56}\)
Hi chetan2u, can you please detail this part ? I'm confused with your calculation here. Why P(not N) is not equal to P(not G) ? Thanks !



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From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
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09 Sep 2016, 12:00
Chance to choose George for one spot: 1/8 Chance to not choose Nina in the other 2 spots = 6/7 * 5/6
Chance that all three will happen: 1/8 * 6/7 * 5/6 = 5/56 Ways to arrange this situation: 3 3 * 5/56 = 15/56
Right?
Edit: I made a small adjustment to my math after realizing that I made a small mistake yet still found the correct answer by luck. The above set up should work.



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Re: From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
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17 Nov 2016, 20:30
Groups including George & not Nina/ Total Groups
Groups including George & not Nina > G __ __ We have 6 people remaining to pick 2 from > 6C2 = 15
Total Groups > 8 people total from which to pick 3 > 8C3 > 56
15/56
C.



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Re: From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
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29 Nov 2016, 16:10
mohitmohit11 wrote: From a Group of 8 People, Including George and Nina, 3 people are to be selected at random to work on a certain project. What is the probability that 3 people selected will include George but not Nina
A 5/56 B 9/56 C 15/56 D 21/56 E 25/56 We are given that 3 people are to be selected from 8 people and need to determine the probability that, of the 3 people selected, George is included but Nina is not. Let’s first determine the total number of ways to select 3 people from a group of 8. The number of ways to select 3 people from a group of 8 is 8C3 = (8 x 7 x 6)/3! = 56. Next, let’s determine the number of ways to select 3 people from a group of 8 when George is included but Nina is not. Since George must be included and Nina is not, we can reduce the number of available spots from 3 to 2 (because George is already selected), and we can reduce the number of people available to be selected from 8 to 6 (Nina is not even considered, and George is already selected). Thus, the number of ways to select 3 people when George is included and Nina is not is: 6C2 = (6 x 5)/2! = 15 So, finally, the probability of selecting a group of three with George and not Nina is: 15/56 Answer: C
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Re: From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
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17 Mar 2018, 10:08
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Dear Friends there are two methods to solve this problem: 1) to count the number of unique 3 people groups (denominator) and the number of 3 people groups that include George but not Nina (numerator) 2) Using probability formula.
Since first method has been clearly explained in previous comments I will focus on the second method
1) 1/8 x 6/7 x 5/6 = 5/56 George selected and Nina is not selected
2) 6/8 x 1/7 x 5/6 = 5/56 First is the probability that neither George nor Nina selected. Why it is 6/8? Because there are 8 candidates, but we want those people who are not George nor Nina. Second is the probability that George is selected. The third is probability that Nina is not selected
3) 6/8 x 5/7 x 1/6 = 5/56 First is the probability that neither George nor Nina will be selected. Second is probability that neither George nor Nina will be selected. Why 5/7? Again because after selecting one person for the first slot, there are 7 people remaining (including George and Nina) We want only 5 of them who are neither George nor Nina
5/56 + 5/56 + 5/56 = 15/56




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