GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Nov 2018, 03:43

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

## Events & Promotions

###### Events & Promotions in November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
• ### All GMAT Club Tests are Free and open on November 22nd in celebration of Thanksgiving Day!

November 22, 2018

November 22, 2018

10:00 PM PST

11:00 PM PST

Mark your calendars - All GMAT Club Tests are free and open November 22nd to celebrate Thanksgiving Day! Access will be available from 0:01 AM to 11:59 PM, Pacific Time (USA)
• ### Free lesson on number properties

November 23, 2018

November 23, 2018

10:00 PM PST

11:00 PM PST

Practice the one most important Quant section - Integer properties, and rapidly improve your skills.

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

Author Message
TAGS:

### Hide Tags

Intern
Joined: 08 Feb 2018
Posts: 1
Anna and Mark are included in a group of 8 people. If 4 people are  [#permalink]

### Show Tags

Updated on: 08 Feb 2018, 19:45
2
00:00

Difficulty:

55% (hard)

Question Stats:

48% (01:46) correct 52% (02:23) wrong based on 47 sessions

### HideShow timer Statistics

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, 12:12.
Last edited by Bunuel on 08 Feb 2018, 19:45, edited 1 time in total.
Renamed the topic, edited the question, moved to PS and added the OA.
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4488
Re: Anna and Mark are included in a group of 8 people. If 4 people are  [#permalink]

### Show Tags

08 Feb 2018, 16:01
4
2
MRLouis wrote:

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 = $$\dfrac{20}{70}$$ = $$\dfrac{2}{7}$$

Does all this make sense?
Mike
_________________

Mike McGarry
Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

##### General Discussion
Math Expert
Joined: 02 Sep 2009
Posts: 50708
Re: Anna and Mark are included in a group of 8 people. If 4 people are  [#permalink]

### Show Tags

08 Feb 2018, 19:52
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.

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 50708
Re: Anna and Mark are included in a group of 8 people. If 4 people are  [#permalink]

### Show Tags

08 Feb 2018, 22:06
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.

Who can find logic behind the following solution:

$$\frac{1 - (\frac{4!}{2!}*\frac{1}{8}*\frac{1}{7}*1*1 + \frac{6}{8}*\frac{5}{7}*\frac{4}{7}*\frac{3}{5})}{2}=\frac{2}{7}$$
_________________
Manager
Joined: 28 Jan 2018
Posts: 53
Location: Netherlands
Concentration: Finance
GMAT 1: 710 Q50 V36
GPA: 3
Re: Anna and Mark are included in a group of 8 people. If 4 people are  [#permalink]

### Show Tags

10 Feb 2018, 12:03
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.

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.
Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2830
Re: Anna and Mark are included in a group of 8 people. If 4 people are  [#permalink]

### Show Tags

12 Feb 2018, 16:36
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.

_________________

Jeffery Miller

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Re: Anna and Mark are included in a group of 8 people. If 4 people are &nbs [#permalink] 12 Feb 2018, 16:36
Display posts from previous: Sort by