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 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
Live EA class + 6 months of EA OnDemand
  • Expert-led weekly online sessions
  • EA Masterclass access between classes
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

130-point score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

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

There were \(r\) red balls and \(y\) yellow balls in a bag.

Expert replies
by AAPL » Sat Aug 24, 2019 10:05 am

Timer

00:00

Answers

A

B

C

D

E

Stats

Difficulty

Economist GMAT

There were \(r\) red balls and \(y\) yellow balls in a bag. Three red balls and four yellow ones were added to the bag. What is the probability that a yellow ball and then a red ball will be selected if Jerry pulls out 2 balls at random, and puts the first ball back before pulling the second ball?

A. \(\frac{y+4}{y+r+7} \cdot \frac{r+3}{y+r+7}\)

B. \(\frac{y+3}{y+r+7} + \frac{r+4}{y+r+6}\)

C. \(\frac{y+3}{y+r+7} \cdot \frac{r+4}{y+r+6}\)

D. \(\frac{y+3}{y+r+7} \cdot \frac{r+3}{y+r+7}\)

E. \(\frac{y+3}{y+r+7} + \frac{r+3}{y+r+6}\)

OA A
Join the discussion
Source: — Problem Solving |

by GMATGuruNY » Sat Aug 24, 2019 12:12 pm
AAPL wrote:Economist GMAT

There were \(r\) red balls and \(y\) yellow balls in a bag. Three red balls and four yellow ones were added to the bag. What is the probability that a yellow ball and then a red ball will be selected if Jerry pulls out 2 balls at random, and puts the first ball back before pulling the second ball?

A. \(\frac{y+4}{y+r+7} \cdot \frac{r+3}{y+r+7}\)

B. \(\frac{y+3}{y+r+7} + \frac{r+4}{y+r+6}\)

C. \(\frac{y+3}{y+r+7} \cdot \frac{r+4}{y+r+6}\)

D. \(\frac{y+3}{y+r+7} \cdot \frac{r+3}{y+r+7}\)

E. \(\frac{y+3}{y+r+7} + \frac{r+3}{y+r+6}\)
Let r=0 and y=0, implying that the bag is EMPTY before the addition of 3 red balls and 4 yellow balls.

P(yellow ball) = 4/7. (Of the 7 balls, 4 are yellow.)
P(red ball) = 3/7. (Of the 7 balls, 3 are red.)
To combine these probabilities, we multiply:
4/7 * 3/7

The correct answer must yield the product above when r=0 and y=0.
Only A works:
\(\frac{y+4}{y+r+7} \cdot \frac{r+3}{y+r+7}\) = \(\frac{0+4}{0+0+7} \cdot \frac{0+3}{0+0+7}\) = \(\frac{4}{7} \cdot \frac{3}{7}\)

The correct answer is A.
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
Join the discussion

by Ian Stewart » Sun Aug 25, 2019 8:18 am
We have y+4 yellow balls and r+3 red balls, and r+y+7 balls in total. So the probability of picking, with replacement, first a yellow then a red ball will be (y+4)/(r+y+7) * (r+3)/(r+y+7), which is answer A.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com
Join the discussion

by Scott@TargetTestPrep » Tue Aug 27, 2019 5:19 pm
AAPL wrote:Economist GMAT

There were \(r\) red balls and \(y\) yellow balls in a bag. Three red balls and four yellow ones were added to the bag. What is the probability that a yellow ball and then a red ball will be selected if Jerry pulls out 2 balls at random, and puts the first ball back before pulling the second ball?

A. \(\frac{y+4}{y+r+7} \cdot \frac{r+3}{y+r+7}\)

B. \(\frac{y+3}{y+r+7} + \frac{r+4}{y+r+6}\)

C. \(\frac{y+3}{y+r+7} \cdot \frac{r+4}{y+r+6}\)

D. \(\frac{y+3}{y+r+7} \cdot \frac{r+3}{y+r+7}\)

E. \(\frac{y+3}{y+r+7} + \frac{r+3}{y+r+6}\)

OA A
The probability that the first ball is yellow is (y + 4)/(y + r + 7). The probability that the second ball is red, after the first ball is put back into the bag, is (r + 3)/(y + r + 7). Therefore, the overall probability is:

(y + 4)/(y + r + 7) * (r + 3)/(y + r + 7)

Answer: A

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
Join the discussion