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Anna and Mark are included in a group of 8 people. If 4 people are

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Anna and Mark are included in a group of 8 people. If 4 people are  [#permalink]

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New post Updated on: 08 Feb 2018, 20:45
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Anna and Mark are included in a group of 8 people. If 4 people are selected at random and without replacement from the group, what is the probability that Mark and not Anna will be selected?

A. 1/7
B. 2/7
C. 3/7
D. 4/7
E. 5/7

Originally posted by MRLouis on 08 Feb 2018, 13:12.
Last edited by Bunuel on 08 Feb 2018, 20:45, edited 1 time in total.
Renamed the topic, edited the question, moved to PS and added the OA.
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Magoosh GMAT Instructor
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Re: Anna and Mark are included in a group of 8 people. If 4 people are  [#permalink]

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New post 08 Feb 2018, 17:01
4
2
MRLouis wrote:
Hello everyone! Can someone please help to solve the question below?

Anna and Mark are included in a group of 8 people. If 4 people are selected at random and without replacemet from thi group, what is the probability tht Mark and not Anna will be selected?

- 1/7
- 2/7
- 3/7
- 4/7
- 5/7


Thank you very mcuh!

Louis

Dear MRLouis,

I'm happy to respond. :-)

This problem can be addressed relatively easily by using probability counting techniques. See:
GMAT Probability and Counting Techniques

Question #1: What's the total number of groups that could be drawn here?

That's given by:

8C4 = \(\dfrac{8!}{(4!)(4!)}\) = \(\dfrac{8*7*6*5}{(4*3*2*1)}\) = 2*7*1*5 = 70

That's the denominator.

Question #2: How many groups would satisfy the condition?

We would have Mark and then could choose 3 from the remaining 6, not including Anna. That's given by:

6C3 = \(\dfrac{6!}{(3!)(3!)}\) = \(\dfrac{6*5*4}{(3*2*1)}\) = 5*4 = 20

That's the numerator of the probability fraction.

Probability = \(\dfrac{20}{70}\) = \(\dfrac{2}{7}\)

Answer = (B)

Does all this make sense?
Mike :-)
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Re: Anna and Mark are included in a group of 8 people. If 4 people are  [#permalink]

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New post 08 Feb 2018, 20:52
2
MRLouis wrote:
Anna and Mark are included in a group of 8 people. If 4 people are selected at random and without replacement from the group, what is the probability that Mark and not Anna will be selected?

A. 1/7
B. 2/7
C. 3/7
D. 4/7
E. 5/7


We need {Mark, Any but Anna, Any but Anna, Any but Anna} in any order.

\(P(MAAA) = \frac{4!}{3!}*\frac{1}{8}*\frac{6}{7}*\frac{5}{6}*\frac{4}{5}=\frac{2}{7}\). We multiply by 4!/3! = 4 because {Mark, Any but Anna, Any but Anna, Any but Anna}, MAAA, can be arranged in four ways: MAAA, AMAA, AAMA, AAAM.

Answer: B.
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Re: Anna and Mark are included in a group of 8 people. If 4 people are  [#permalink]

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New post 08 Feb 2018, 23:06
1
Bunuel wrote:
MRLouis wrote:
Anna and Mark are included in a group of 8 people. If 4 people are selected at random and without replacement from the group, what is the probability that Mark and not Anna will be selected?

A. 1/7
B. 2/7
C. 3/7
D. 4/7
E. 5/7


We need {Mark, Any but Anna, Any but Anna, Any but Anna} in any order.

\(P(MAAA) = \frac{4!}{3!}*\frac{1}{8}*\frac{6}{7}*\frac{5}{6}*\frac{4}{5}=\frac{2}{7}\). We multiply by 4!/3! = 4 because {Mark, Any but Anna, Any but Anna, Any but Anna}, MAAA, can be arranged in four ways: MAAA, AMAA, AAMA, AAAM.

Answer: B.


Who can find logic behind the following solution:

\(\frac{1 - (\frac{4!}{2!}*\frac{1}{8}*\frac{1}{7}*1*1 + \frac{6}{8}*\frac{5}{7}*\frac{4}{7}*\frac{3}{5})}{2}=\frac{2}{7}\)
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Re: Anna and Mark are included in a group of 8 people. If 4 people are  [#permalink]

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New post 10 Feb 2018, 13:03
1
Bunuel wrote:
Bunuel wrote:
MRLouis wrote:
Anna and Mark are included in a group of 8 people. If 4 people are selected at random and without replacement from the group, what is the probability that Mark and not Anna will be selected?

A. 1/7
B. 2/7
C. 3/7
D. 4/7
E. 5/7


We need {Mark, Any but Anna, Any but Anna, Any but Anna} in any order.

\(P(MAAA) = \frac{4!}{3!}*\frac{1}{8}*\frac{6}{7}*\frac{5}{6}*\frac{4}{5}=\frac{2}{7}\). We multiply by 4!/3! = 4 because {Mark, Any but Anna, Any but Anna, Any but Anna}, MAAA, can be arranged in four ways: MAAA, AMAA, AAMA, AAAM.

Answer: B.




Who can find logic behind the following solution:

\(\frac{1 - (\frac{4!}{2!}*\frac{1}{8}*\frac{1}{7}*1*1 + \frac{6}{8}*\frac{5}{7}*\frac{4}{7}*\frac{3}{5})}{2}=\frac{2}{7}\)


The numerator of the left equation can be understand as: 1 - ( probability that both are selected + probability both are not selected) and it's equal to (only 1 of them is selected), we need to divide it by 2 because the numerator is account for both situation, where only Anna is chosen or only Mark is chosen, and we only want Mark.
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Re: Anna and Mark are included in a group of 8 people. If 4 people are  [#permalink]

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New post 12 Feb 2018, 17:36
MRLouis wrote:
Anna and Mark are included in a group of 8 people. If 4 people are selected at random and without replacement from the group, what is the probability that Mark and not Anna will be selected?

A. 1/7
B. 2/7
C. 3/7
D. 4/7
E. 5/7


The calculation of the number of ways to select Mark but not Anna requires that we only consider the 6 remaining individuals because we don’t count Anna at all, and we already know that Mark has been selected. Thus, there are 6 available individuals to fill just 3 slots: 6C3 = 6!/[3!(6-3)!] = (6 x 5 x 4)/(3 x 2 x 1) = 20.

The number of ways to select 4 people from 8 is 8C4 = 8!/[4!(8-4)!] = (8 x 7 x 6 x 5)/(4 x 3 x 2 x 1) = 7 x 2 x 5 = 70.

So the probability that Mark will be selected and Anna will not be selected is 20/70 = 2/7.

Answer: B
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Re: Anna and Mark are included in a group of 8 people. If 4 people are &nbs [#permalink] 12 Feb 2018, 17:36
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