From a Group of 8 People, Including George and Nina, 3 people are

Question

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

OptimusPrepJanielle · Accepted Answer

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

Kurtosis · Answer

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

Bunuel · Answer

Similar question to practice: from-a-group-of-8-volunteers-including-andrew-and-karen-160883.html

GSBae · Answer

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

Alex75PAris · Answer

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}

iliavko · Answer

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

G-George N-nina   We want to know  George only , so no Nina and no overlap George plus Nina.

P(N)=1-P(not N)    1- \frac{7}{8}*\frac{6}{7}*\frac{5}{6} = \frac{3}{8} 
 P(G)=1-P(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: 1-P(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!

Alex75PAris · Answer

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 occurs 
Stage 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 outcomes 
Number of ways choose 3 people from 8 = 56

So P = 15/56

OK ?

iliavko · Answer

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.

chetan2u · Answer

Hi,

You are going wrong in execution of your DATA..

P(N)=1-P(not N)    1- \frac{7}{8}*\frac{6}{7}*\frac{5}{6} = \frac{3}{8} 
 P(G)=1-P(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{24-9}{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..

iliavko · Answer

Hi Chetan!

Cool!, Thank you so much for the clarification!

So I guess the way I did it I was double-counting 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!

Alex75PAris · Answer

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) * = 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 !!

ScottTargetTestPrep · Answer

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

kablayi · Answer

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

Deepak23gmat · Answer

Can we use fundamental counting principle to achieve the no of ways for selecting 3 persons from 8 with George included & Nina excluded.  Bunuel   KarishmaB

I used same & got 6x5 = 30 ways, please help me figure out the gap/flaw in my approach.

Thanks in advance
Deepak

Bunuel · Answer

6 * 5 counts both X-Y and Y-X pairs as distinct, even though the order doesn’t matter here. You should divide that number by 2 to avoid double-counting.

KarishmaB · Answer

FCP arranges the outcomes for you which means that if you want to just "select" you must un-arrange. As Bunuel suggested, you need to divide by 2 to un-arrange (just like to arrange, you would multiply by 2) Check this video for discussion on FCP: https://youtu.be/LFnLKx06EMU

egmat · Answer

This is a probability problem with constraints - these can definitely be tricky until you see the underlying pattern. Let me walk you through how to think about this one.

Understanding What We're Asked

You need to find the probability that when selecting 3 people from 8, George is included but Nina is not. Let's break this down step by step.

The Key Insight

Here's what you need to see: if George  must  be selected, then he's automatically taking up one of the three spots. And if Nina  cannot  be selected, she's completely out of consideration.

Think about it this way - you've already filled 1 spot (with George), and you've eliminated Nina from the pool. So really, you're just choosing 2 more people from the remaining 6 people (8 total minus George and Nina).

Let's Count the Favorable Outcomes

Since George is already selected and Nina is out, we need to choose 2 people from the 6 remaining people.

Number of ways =  C(6,2) = \frac{6!}{2! 	imes 4!} = \frac{6 	imes 5}{2 	imes 1} = 15

Total Possible Outcomes

Without any restrictions, the total number of ways to select 3 people from 8 is:

C(8,3) = \frac{8!}{3! 	imes 5!} = \frac{8 	imes 7 	imes 6}{3 	imes 2 	imes 1} = \frac{336}{6} = 56

Calculate the Probability

Probability =  \frac{	ext{Favorable outcomes}}{	ext{Total outcomes}} = \frac{15}{56}

The answer is  C: 15/56

Notice how the constraint actually simplified our problem - instead of dealing with complex conditional probabilities, we just adjusted our selection space. This makes intuitive sense too: the probability (about 0.27) is less than 1/2, which it should be since we have restrictive conditions.

For the complete framework that applies to all constraint-based selection problems, plus alternative approaches and time-saving techniques, you can check out the  step-by-step solution on Neuron by e-GMAT . You'll discover the systematic method to handle any "must include/must exclude" variation quickly. You can also explore other GMAT official questions with detailed solutions on Neuron for structured practice  here .

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