It is currently 20 Jul 2017, 17:55

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Sam and Jessica are invited to a dance. If there are 7 men and 7 women

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

1 KUDOS received
Intern
Intern
avatar
Joined: 15 Oct 2009
Posts: 12
Sam and Jessica are invited to a dance. If there are 7 men and 7 women [#permalink]

Show Tags

New post 18 Mar 2010, 23:56
1
This post received
KUDOS
1
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

62% (01:47) correct 38% (01:24) wrong based on 27 sessions

HideShow timer Statistics

Sam and Jessica are invited to a dance. If there are 7 men and 7 women in total at the dance, and one woman and one man are chosen to lead the dance, what is the probability that Sam and Jessica will NOT be the pair chosen to lead the dance?

A) 1/49
B) 1/7
C) 6/7
D) 47/49
E) 48/49

[Reveal] Spoiler:
I came up with the answer but I tried to solve it in a different way originally and failed miserably.

Why can't i come up with the right fraction if I say, there are 6/7 chance that Sam will NOT be picked and 6/7 chance that Jessican will NOT be picked.
So the probably is 36/49 that Sam and Jessica will NOT be the pair choosen. Why wouldn't this method work?
[Reveal] Spoiler: OA
Expert Post
2 KUDOS received
Math Expert
User avatar
D
Joined: 02 Sep 2009
Posts: 40281
Re: Sam and Jessica are invited to a dance. If there are 7 men and 7 women [#permalink]

Show Tags

New post 19 Mar 2010, 05:20
2
This post received
KUDOS
Expert's post
changhiskhan wrote:
Sam and Jessica are invited to a dance. If there are 7 men and 7 women in total at the dance, and one woman and one man are chosen to lead the dance, what is the probability that Sam and Jessica will NOT be the pair chosen to lead the dance?

A) 1/49 B) 1/7 C) 6/7 D) 47/49 E) 48/49

I came up with the answer but I tried to solve it in a different way originally and failed miserably.

Why can't i come up with the right fraction if I say, there are 6/7 chance that Sam will NOT be picked and 6/7 chance that Jessican will NOT be picked.
So the probably is 36/49 that Sam and Jessica will NOT be the pair choosen. Why wouldn't this method work?


Welcome to the Gmat Club.

The easiest way to solve this problem is to calculate opposite probability and subtract this value from 1:
The opposite probability is that Sam and Jessica WILL be chosen to lead \(=\frac{1}{7}*\frac{1}{7}=\frac{1}{49}\).
So, the probability that they will NOT be chosen to lead \(= 1-\frac{1}{49}=\frac{48}{49}\).

Answer: E.

The way you are doing is wrong because \(\frac{36}{49}\) is the probability that in the leading pair won't be neither Sam nor Jessica. But in the leading pair can be Sam with any woman but Jessica and can be Jessica with any man but Sam.

So, if you are doing this way you should count \(\frac{7}{7}*\frac{6}{7}\) probability of pairing any man (\(\frac{7}{7}\)) with any woman but Jessica (\(\frac{6}{7}\)) PLUS \(\frac{1}{7}*\frac{6}{7}\) probability of pairing Jessica (\(\frac{1}{7}\)) with any man but Sam (\(\frac{6}{7}\)) \(=\frac{7}{7}*\frac{6}{7}+\frac{1}{7}*\frac{6}{7}=\frac{48}{49}\).

OR you can think about it this way: there are total \(7*7=49\) pairs possible, and only one pair is when Jessica and Sam are together, hence probability \(=\frac{49-1}{49}=\frac{48}{49}\)

OR: pairs without Jessica \(6*7=42\) and pairs with Jessica but without Sam \(1*6=6\), hence probability \(=\frac{42+6}{49}=\frac{48}{49}\).

Hope it helps.
_________________

New to the Math Forum?
Please read this: All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Intern
Intern
avatar
Joined: 19 Mar 2010
Posts: 11
Location: San Diego, CA
Schools: UCLA Anderson
Re: Sam and Jessica are invited to a dance. If there are 7 men and 7 women [#permalink]

Show Tags

New post 19 Mar 2010, 09:12
I agree with the previous poster. For probability, the basic equation is:

# of ways event can happen / total # of possible ways

The numbers of pairs that have Sam and Jessica is just 1, out of a total possibility of 49 pairs. So 1/49 chance they will be chosen.

The chance they will NOT be chosen is 1 minus this value, or 1- 1/49 = 48/49.
Intern
Intern
avatar
Joined: 15 Oct 2009
Posts: 12
Re: Sam and Jessica are invited to a dance. If there are 7 men and 7 women [#permalink]

Show Tags

New post 19 Mar 2010, 09:19
Thank you for the wonderful explanation.
Senior Manager
Senior Manager
User avatar
Joined: 13 Dec 2009
Posts: 262
Reviews Badge
Re: Sam and Jessica are invited to a dance. If there are 7 men and 7 women [#permalink]

Show Tags

New post 19 Mar 2010, 12:51
Good explanation. And +1 for E 48/49
_________________

My debrief: done-and-dusted-730-q49-v40

Intern
Intern
avatar
Joined: 26 Jul 2010
Posts: 5
probabilty problem "Sam & Jessica" [#permalink]

Show Tags

New post 26 Jul 2010, 23:27
Hi!
Could one of the moderators please help me with it?
"Sam and Jessica are invited to a dance. If there are 7 men and 7 women in total at the dance, and one woman and one man ar chosen to lead the dance, what is the probability that Sam and Jessica will NOT be chosen to lead the dance?"

I know, the thread was posted a looooooonnnnnnng time ago, but I am so confused with which method I should take and WHY?

My question is: why can't I solve the problem in this way?
1 - C 1 7/C2 14= 12/13
C1 7 : I would choose one group out of 7 (since there are 14 people)
C2 14: altogether I would have chosen 2 people out of 14

The "funny" point is, I do also understand, how the original answer has been calculated.
Can someone please help me out?

Many thanks!!!!

OA:
[Reveal] Spoiler:
48/49
Expert Post
CEO
CEO
User avatar
Joined: 17 Nov 2007
Posts: 3583
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
GMAT ToolKit User Premium Member
Re: probabilty problem "Sam & Jessica" [#permalink]

Show Tags

New post 29 Jul 2010, 10:35
1. You use p = 1 - q formula in which q means "Sam and Jessica"

2. \(C^{14}_2\) as the total number of all possibilities are incorrect as it counts man-man and woman-woman pairs. So, you have to use \(C^7_1 * C^7_1\) = 49 or any man and any woman combinations.

3. \(C^7_1\). I don't get the reasoning behind the formula. There is only one way to choose "Sam and Jessica" and why do you think that there is 7 groups? There is no special assignment for each woman and man.

Let me know if you need further help.
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Expert Post
CEO
CEO
User avatar
Joined: 17 Nov 2007
Posts: 3583
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
GMAT ToolKit User Premium Member
Re: probabilty problem "Sam & Jessica" [#permalink]

Show Tags

New post 29 Jul 2010, 10:47
By the way, welcome to GMAT Club! :)
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Intern
Intern
avatar
Joined: 26 Jul 2010
Posts: 5
Re: probabilty problem "Sam & Jessica" [#permalink]

Show Tags

New post 31 Jul 2010, 10:10
walker wrote:
1. You use p = 1 - q formula in which q means "Sam and Jessica"

2. \(C^{14}_2\) as the total number of all possibilities are incorrect as it counts man-man and woman-woman pairs. So, you have to use \(C^7_1 * C^7_1\) = 49 or any man and any woman combinations.

3. \(C^7_1\). I don't get the reasoning behind the formula. There is only one way to choose "Sam and Jessica" and why do you think that there is 7 groups? There is no special assignment for each woman and man.

Let me know if you need further help.


Hi, "walker"!

Thank you for your explanation!

Could you be so kind and explain to me, when to use what kind of formula? My problem is, I often mix combinatorics-probability formula with simple probability formula. Is there a helpful strategy?

Again, thank you so much!
Expert Post
CEO
CEO
User avatar
Joined: 17 Nov 2007
Posts: 3583
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
GMAT ToolKit User Premium Member
Re: probabilty problem "Sam & Jessica" [#permalink]

Show Tags

New post 01 Aug 2010, 04:36
Hi fledgling2010,

I think I understand why GMAC use "counting" rather than "prob/comb" term. It seems you are trying to remember where which formula to use rather than understand how to "count". So, my advice is to read this [http://gmatclub.com/forum/math-probability-87244.html][http://gmatclub.com/forum/math-combinatorics-87345.html] and each time you see prob/comb problem try to get idea how to count and then use an appropriate formula.
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Senior Manager
Senior Manager
avatar
Joined: 03 Nov 2005
Posts: 385
Location: Chicago, IL
Re: probabilty problem "Sam & Jessica" [#permalink]

Show Tags

New post 02 Aug 2010, 17:44
I personally think this problem is solved so much easier using simple probability rules

The trick here is to estimate the individual probabilities of selecting him and her and multiply them because the events are independent from each other. the probability of selecting 1 individual out of 7 is 1/7 and the probability of selecting him and her is 1/7*1/7=1/49.

P of not selecting them both is what is asked for.

1-1/49 = 48/49.

I hope u find this method much more clear and concise.
_________________

Hard work is the main determinant of success

GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 09 Sep 2013
Posts: 16468
Premium Member
Re: Sam and Jessica are invited to a dance. If there are 7 men and 7 women [#permalink]

Show Tags

New post 26 Nov 2016, 03:07
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Re: Sam and Jessica are invited to a dance. If there are 7 men and 7 women   [#permalink] 26 Nov 2016, 03:07
    Similar topics Author Replies Last post
Similar
Topics:
1 Experts publish their posts in the topic John can mow 7 acres of land in one day, while Sam and John can mow 22 Bunuel 3 19 Apr 2017, 16:04
5 Experts publish their posts in the topic From a group of 7 men and 6 women, five persons are to be selected to azamaka 6 27 Jan 2017, 20:56
5 Experts publish their posts in the topic 70% of stadium visitors are men. 1/7 of men and 1/5 of women at the st Bunuel 6 24 Jul 2015, 08:33
7 Experts publish their posts in the topic Initially, the men and women in a room were in the ratio of 5:7. Six Bunuel 12 13 May 2017, 09:20
Experts publish their posts in the topic (7^6) + (7^8) = mikemcgarry 1 19 Sep 2012, 15:55
Display posts from previous: Sort by

Sam and Jessica are invited to a dance. If there are 7 men and 7 women

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.