BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions
Start free 5-day trial
Read 559 reviewsTarget Test Prep

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Ashley and Vinnie work on a sales staff with 8 other

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Moderator
Posts: 7187
Joined: Thu Sep 07, 2017 4:43 pmFollowed by: 23 members
by BTGmoderatorDC » Wed Mar 13, 2019 8:51 pm

Timer

00:00

Answers

A

B

C

D

E

Stats

Difficulty—

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

OA C

Source: Veritas Prep
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3008
Joined: Mon Aug 22, 2016 6:19 amFollowed by: 34 members
Location: Grand Central / New York
Thanked: 470 times

by Jay@ManhattanReview » Wed Mar 13, 2019 9:04 pm
BTGmoderatorDC 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

OA C

Source: Veritas Prep
Including Ashley and Vinnie, there are 10 salespeople.

Number of ways of choosing ANY 5 members out of 10 = 10C5;
Number of ways of choosing 5 members out of 10 such that 2 of them are Ashley and Vinnie = 2C2 * 8C3 = 8C3;

Probability that both Ashley and Vinnie will be chosen = 8C3 / 10C5 = 2/9

The correct answer: C

Hope this helps!

-Jay
_________________
Manhattan Review GMAT Prep

Locations: GMAT Classes Glasgow | GMAT Prep Courses Hyderabad | LSAT Prep Courses San Diego | Manhattan Prep Classes SAT | and many more...

Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members
Location: New York, NY
Thanked: 13060 times
GMAT Score:790

by GMATGuruNY » Thu Mar 14, 2019 1:59 am
BTGmoderatorDC 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
From the 10 people, 5 will be selected.
Thus, P(Ashley is selected) = 5/10.
From the 9 remaining people, 4 will be selected.
Thus, P(Vinnie is selected) = 4/9.
To combine these probabilities, we multiply:
5/10 * 4/9 = 2/9.

The correct answer is C.

Similar problem:
https://www.beatthegmat.com/probability ... 72268.html
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 8210
Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members
Location: Los Angeles, CA
Thanked: 43 times

by Scott@TargetTestPrep » Mon Mar 18, 2019 5:34 pm
BTGmoderatorDC 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

OA C

Source: Veritas Prep
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
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

ImageImage
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1