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

 It is currently 20 Apr 2019, 01:29

### 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

# Ashley and Vinnie work on a sales staff with 8 other salespeople. If f

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

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 54375
Ashley and Vinnie work on a sales staff with 8 other salespeople. If f  [#permalink]

### Show Tags

21 Apr 2015, 06:07
3
12
00:00

Difficulty:

45% (medium)

Question Stats:

68% (02:34) correct 32% (02:20) wrong based on 168 sessions

### HideShow timer Statistics

Ashley and Vinnie work on a sales staff with 8 other salespeople. If five members of the staff will be chosen at random to attend a sales workshop, what is the probability that both Ashley and Vinnie will be chosen?

A. 1/10
B. 3/28
C. 2/9
D. 1/4
E. 1/2

Kudos for a correct solution.

_________________
##### Most Helpful Community Reply
Senior Manager
Joined: 02 Mar 2012
Posts: 293
Schools: Schulich '16
Re: Ashley and Vinnie work on a sales staff with 8 other salespeople. If f  [#permalink]

### Show Tags

21 Apr 2015, 21:00
4
1
C -2/9

have to chose 2 members and 3 more out of 10

so 2c2*8c3/10c5

simple..which gives 2/9

hope it helps
##### General Discussion
Retired Moderator
Joined: 06 Jul 2014
Posts: 1228
Location: Ukraine
Concentration: Entrepreneurship, Technology
GMAT 1: 660 Q48 V33
GMAT 2: 740 Q50 V40
Ashley and Vinnie work on a sales staff with 8 other salespeople. If f  [#permalink]

### Show Tags

Updated on: 21 Apr 2015, 11:20
update: wrong explanation

Bunuel wrote:
Ashley and Vinnie work on a sales staff with 8 other salespeople. If five members of the staff will be chosen at random to attend a sales workshop, what is the probability that both Ashley and Vinnie will be chosen?

A. 1/10
B. 3/28
C. 2/9
D. 1/4
E. 1/2

Kudos for a correct solution.

Let's imagine that Ashley and Vinnie this is one person and in total we have 9 salespersons. We have 4C9 = 126 possible combinations and in half of this combinations will be present Ashley+Vinnie.
So we have 63 possible combinations when Ashley and Vinnie chosen.

And in real situtaion we have 5C10 = 252 possible combinations. And as we already know in 63 combinations will be present Ashley and Vinnie.
So probability equal: $$\frac{63}{252} = \frac{1}{4}$$

Answer is D

_________________

Originally posted by Harley1980 on 21 Apr 2015, 09:18.
Last edited by Harley1980 on 21 Apr 2015, 11:20, edited 1 time in total.
Manager
Joined: 27 Dec 2013
Posts: 222
Re: Ashley and Vinnie work on a sales staff with 8 other salespeople. If f  [#permalink]

### Show Tags

21 Apr 2015, 10:22
1
Hi Harley1980.

I did slightly different and hence obviously ended up with the answer 2/9.

I assumed that because Ashley and Winni have to be chosen, the remaining selection is 3 people from 8.

8C3/10C5= 2/9. Option C.

Please could you explain your answer.

Harley1980 wrote:
Bunuel wrote:
Ashley and Vinnie work on a sales staff with 8 other salespeople. If five members of the staff will be chosen at random to attend a sales workshop, what is the probability that both Ashley and Vinnie will be chosen?

A. 1/10
B. 3/28
C. 2/9
D. 1/4
E. 1/2

Kudos for a correct solution.

Let's imagine that Ashley and Vinnie this is one person and in total we have 9 salespersons. We have 5C9 = 126 possible combinations and in half of this combinations will be present Ashley+Vinnie.
So we have 63 possible combinations when Ashley and Vinnie chosen.

And in real situtaion we have 5C10 = 252 possible combinations. And as we already know in 63 combinations will be present Ashley and Vinnie.
So probability equal: $$\frac{63}{252} = \frac{1}{4}$$

_________________
Kudos to you, for helping me with some KUDOS.
Retired Moderator
Joined: 06 Jul 2014
Posts: 1228
Location: Ukraine
Concentration: Entrepreneurship, Technology
GMAT 1: 660 Q48 V33
GMAT 2: 740 Q50 V40
Ashley and Vinnie work on a sales staff with 8 other salespeople. If f  [#permalink]

### Show Tags

21 Apr 2015, 11:19
update: wrong explanation
shriramvelamuri wrote:
Hi Harley1980.

I did slightly different and hence obviously ended up with the answer 2/9.

I assumed that because Ashley and Winni have to be chosen, the remaining selection is 3 people from 8.

8C3/10C5= 2/9. Option C.

Please could you explain your answer.

Harley1980 wrote:
Bunuel wrote:
Ashley and Vinnie work on a sales staff with 8 other salespeople. If five members of the staff will be chosen at random to attend a sales workshop, what is the probability that both Ashley and Vinnie will be chosen?

A. 1/10
B. 3/28
C. 2/9
D. 1/4
E. 1/2

Kudos for a correct solution.

Let's imagine that Ashley and Vinnie this is one person and in total we have 9 salespersons. We have 5C9 = 126 possible combinations and in half of this combinations will be present Ashley+Vinnie.
So we have 63 possible combinations when Ashley and Vinnie chosen.

And in real situtaion we have 5C10 = 252 possible combinations. And as we already know in 63 combinations will be present Ashley and Vinnie.
So probability equal: $$\frac{63}{252} = \frac{1}{4}$$

I'll be honest: I'm not quite sure in my answer

But I think that I see mistake in your calculations
You wrote: "I assumed that because Ashley and Winni have to be chosen, the remaining selection is 3 people from 8. "
But I think we can't do this, for example we need to calculate 4C6 = 15 variants
and if we presuppose that first position is taken and calculate remained 3 position from 5 variants: 3C5 we receive 10 variants + 1 number, but this is wrong approach because we lost 5 variants.

_________________
EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 13956
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: Ashley and Vinnie work on a sales staff with 8 other salespeople. If f  [#permalink]

### Show Tags

21 Apr 2015, 19:53
2
Hi shriramvelamuri,

The 'math' behind this question can be done in a couple of different ways; your approach IS one of them.

Hi Harley1980,

I understand the logic that you used, but by 'combining' Ashley and Vinnie into '1 person', you've changed the parameters of the question (and the 'inputs' into the equations). Since there are 10 people in total (not 9), you have to account for the possible outcomes that would include only 1 of the 2 people (just Ashley or just Vinnie) and that doesn't happen in your calculations. Those options decrease the probability that BOTH Ashley and Vinnie will be chosen, which is why the correct answer is LESS than 1/4.

GMAT assassins aren't born, they're made
Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****

# Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save $75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ Manager Joined: 27 Dec 2013 Posts: 222 Re: Ashley and Vinnie work on a sales staff with 8 other salespeople. If f [#permalink] ### Show Tags 21 Apr 2015, 20:19 Hi Empower, Can we then safely assume the answer is 2/9. EMPOWERgmatRichC wrote: Hi shriramvelamuri, The 'math' behind this question can be done in a couple of different ways; your approach IS one of them. Hi Harley1980, I understand the logic that you used, but by 'combining' Ashley and Vinnie into '1 person', you've changed the parameters of the question (and the 'inputs' into the equations). Since there are 10 people in total (not 9), you have to account for the possible outcomes that would include only 1 of the 2 people (just Ashley or just Vinnie) and that doesn't happen in your calculations. Those options decrease the probability that BOTH Ashley and Vinnie will be chosen, which is why the correct answer is LESS than 1/4. GMAT assassins aren't born, they're made Rich _________________ Kudos to you, for helping me with some KUDOS. EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 13956 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: Ashley and Vinnie work on a sales staff with 8 other salespeople. If f [#permalink] ### Show Tags 21 Apr 2015, 21:30 1 Hi shriramvelamuri, Yes, your approach and explanation are correct. GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com *****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***** # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/
Current Student
Joined: 25 Nov 2014
Posts: 98
Concentration: Entrepreneurship, Technology
GMAT 1: 680 Q47 V38
GPA: 4
Re: Ashley and Vinnie work on a sales staff with 8 other salespeople. If f  [#permalink]

### Show Tags

22 Apr 2015, 00:24
2
Favourable cases : Selecting Ashley and Vinnie in 1 way, and then selecting 3 more people from remaining 8.
thus , Fav = 1 * 8C3.
Total cases = 10C5
Thus prob : 8C3 / 10C5 = 2/9
Ans C.
_________________
Kudos!!
Math Expert
Joined: 02 Sep 2009
Posts: 54375
Re: Ashley and Vinnie work on a sales staff with 8 other salespeople. If f  [#permalink]

### Show Tags

27 Apr 2015, 03:05
Bunuel wrote:
Ashley and Vinnie work on a sales staff with 8 other salespeople. If five members of the staff will be chosen at random to attend a sales workshop, what is the probability that both Ashley and Vinnie will be chosen?

A. 1/10
B. 3/28
C. 2/9
D. 1/4
E. 1/2

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:

This problem includes a mix of combinatorics and probability. The probability that both Ashley and Vinnie will be chosen is: the number of teams on which they're both chosen divided by the number of total possible teams. In these situations it's usually easier to start with the total. With 10 total people and 5 chosen, the number of combinations (order doesn't matter!) can be calculated as:

10!/(5!5!), which factors out to (10)(9)(8)(7)(6)/[(5)(4)(3)(2)]. That ultimately reduces to 252, but remember that you're doing all of this to find a denominator, the number of total teams. So you may want to leave the math undone; as you factor the numerator 6 with the denominator 2 and 3, and then divide the 10 by 5 and the 8 by 4, you're really left with: 2 * 9 * 2 * 7, which you'll find makes an easier denominator. Then for the number of teams that both Vinnie and Ashley are on. The logic here: if those two are on the team, then there are 8 people left to be chosen for 3 remaining spots. So the calculation is: 8!/(3!5!), which reduces to (8)(7)(6)/[(3)(2)], which is 56 (or you can just leave the 8 * 7 undone). That leaves you with a "teams they're on, divided by teams total" calculation of: (8)(7)/[(2)(2)(9)(7)], in which the 7s cancel and the 8 divided by (2 * 2) leaves a 2 in the numerator. That simplifies to 2/9.
_________________
Intern
Joined: 12 Oct 2014
Posts: 2
Re: Ashley and Vinnie work on a sales staff with 8 other salespeople. If f  [#permalink]

### Show Tags

01 Jul 2015, 14:01
[2C2*8C3]/[10C5]=2/9
Manager
Joined: 29 Nov 2016
Posts: 108
Re: Ashley and Vinnie work on a sales staff with 8 other salespeople. If f  [#permalink]

### Show Tags

19 Feb 2018, 23:49
Hi. Can any one suggest how to do this by probability method?

Posted from my mobile device

Posted from my mobile device
Intern
Joined: 29 Apr 2017
Posts: 26
Location: India
Concentration: Operations, Other
GMAT 1: 660 Q43 V38
GPA: 4
WE: Operations (Transportation)
Re: Ashley and Vinnie work on a sales staff with 8 other salespeople. If f  [#permalink]

### Show Tags

18 Nov 2018, 01:08
Total Outcome = 10C5 (selecting 5 out of 10)
favourable outcome = 8C3 (selecting 3 out of 8 people as 2 are already selected- Ashley & Vinnie)
Probability = favourable outcome / total outcome = 8C3 / 10C5 = 2/9
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 5807
Location: United States (CA)
Re: Ashley and Vinnie work on a sales staff with 8 other salespeople. If f  [#permalink]

### Show Tags

18 Mar 2019, 18:50
Bunuel wrote:
Ashley and Vinnie work on a sales staff with 8 other salespeople. If five members of the staff will be chosen at random to attend a sales workshop, what is the probability that both Ashley and Vinnie will be chosen?

A. 1/10
B. 3/28
C. 2/9
D. 1/4
E. 1/2

Kudos for a correct solution.

There are 10C5 = 10!/(5! x 5!) = (10 x 9 x 8 x 7 x 6)/(5 x 4 x 3 x 2) = 2 x 9 x 2 x 7 = 252 ways to choose 5 people from 10. Assuming Ashley and Vinnie are already chosen, there are only 3 spots left for the other 8 people, and the number of ways to choose 3 people from 8 is 8C3 = 8!/(3! x 5!) = (8 x 7 x 6)/(3 x 2) = 56. Therefore, the probability is 56/252 = 8/36 = 2/9.

Answer: C
_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

Re: Ashley and Vinnie work on a sales staff with 8 other salespeople. If f   [#permalink] 18 Mar 2019, 18:50
Display posts from previous: Sort by

# Ashley and Vinnie work on a sales staff with 8 other salespeople. If f

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

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne 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®.