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
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
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.

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.

A box contains 10 balls numbered from 1 to 10 inclusive. If

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 » Fri Nov 09, 2018 12:04 am

Timer

00:00

Answers

A

B

C

D

E

Stats

Difficulty—

A box contains 10 balls numbered from 1 to 10 inclusive. If Ann removes a ball at random and replaces it, and then Jane removes a ball at random, what is the probability that both women removed the same ball?

A. 1/100
B. 1/90
C. 1/45
D. 1/10
E. 41/45

OA D

Source: Magoosh
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
User avatar
GMAT Instructor
Posts: 1449
Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members
Thanked: 59 times

A box contains 10 balls numbered from 1 to 10 inclusive. If

by fskilnik@GMATH » Fri Nov 09, 2018 5:13 am
BTGmoderatorDC wrote:A box contains 10 balls numbered from 1 to 10 inclusive. If Ann removes a ball at random and replaces it, and then Jane removes a ball at random, what is the probability that both women removed the same ball?

A. 1/100
B. 1/90
C. 1/45
D. 1/10
E. 41/45
Source: Magoosh
$$?\,\, = \,\,P\left( {{\rm{same}}\,\,{\rm{ball}}\,\,{\rm{twice}}\,{\rm{,}}\,{\rm{with}}\,\,{\rm{replacement}}} \right)$$
Exactly 1 ball extracted in the second extraction (out of 10 equiprobable possibilities) will be the same as (any given) ball extracted in the first extraction.

Therefore the answer is 1/10, immediately.


Another approach:
$${\rm{Total}} = 10 \cdot 10\,\,{\rm{pairs}}\,\,\,\left( {{1^{{\rm{st}}}}\,,\,\,{2^{{\rm{nd}}}}} \right)\,{\rm{ extracted}}\,,\,\,{\rm{all}}\,\,{\rm{equiprobable}}$$
$${\rm{Favorable}}\,\,{\rm{ = }}\,\,{\rm{10}} \cdot {\rm{1}}\,\,{\rm{pairs}}\,{\rm{among}}\,{\rm{the}}\,{\rm{above}}\,\,\,\,\,\left[ {\,\left( {{1^{{\rm{st}}}} = {\rm{any}}\,\,{\rm{ball}}\,,\,\,{2^{{\rm{nd}}}} = {\rm{same}}\,\,{1^{{\rm{st}}}}\,\,{\rm{ball}}} \right)\,} \right]\,$$
$$? = {{10} \over {100}} = {1 \over {10}}$$

These solutions follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 416
Joined: Thu Oct 15, 2009 11:52 am
Thanked: 27 times

by regor60 » Fri Nov 09, 2018 6:30 am
BTGmoderatorDC wrote:A box contains 10 balls numbered from 1 to 10 inclusive. If Ann removes a ball at random and replaces it, and then Jane removes a ball at random, what is the probability that both women removed the same ball?

A. 1/100
B. 1/90
C. 1/45
D. 1/10
E. 41/45

OA D

Source: Magoosh
How many ways to take a distinct pair of numbers that don't match out of 10 numbers ? 10!/2!8! = 45. This is like without replacement in order to avoid a matching pair.

How many ways to take a distinct pair out of 10 numbers with replacement ? 10^2/2 = 50. This includes the above plus matching pairs.

To find the number of matching pairs, subtract the first answer from the second: 50-45 =5

Probability of selecting a matching pair is then 5/50 = [spoiler]D,1/10[/spoiler]
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members
Location: Vancouver, BC
Thanked: 5254 times
GMAT Score:770

by Brent@GMATPrepNow » Sat Nov 10, 2018 8:13 am
BTGmoderatorDC wrote:A box contains 10 balls numbered from 1 to 10 inclusive. If Ann removes a ball at random and replaces it, and then Jane removes a ball at random, what is the probability that both women removed the same ball?

A. 1/100
B. 1/90
C. 1/45
D. 1/10
E. 41/45
P(Ann and Jane remove same ball) = P(Ann removes ANY ball AND Jane's ball matches Ann's ball)
= P(Ann removes ANY ball) x P(Jane's ball matches Ann's ball)
= 1 x 1/10
= 1/10

Answer: D

Aside: Once Jane removes her ball (and then replaces it), we have 10 balls, and 1 of them is the one that Jane picked.
So, P(Jane's ball matches Ann's ball) = 1/10

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 8188
Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members
Location: Los Angeles, CA
Thanked: 43 times

by Scott@TargetTestPrep » Mon Nov 12, 2018 7:18 am
BTGmoderatorDC wrote:A box contains 10 balls numbered from 1 to 10 inclusive. If Ann removes a ball at random and replaces it, and then Jane removes a ball at random, what is the probability that both women removed the same ball?

A. 1/100
B. 1/90
C. 1/45
D. 1/10
E. 41/45
Since Ann can pick any ball and Jane has only a 1/10 chance to match it, the probability that they select the same ball is:

1 x 1/10 = 1/10

Alternate Solution:

Many students think that the answer is 1/10 x 1/10 = 1/100, thinking that each woman has a 1/10 probability of picking a particular ball. Let's look at the "long" solution to this problem to clarify the correct answer of 1/10.

Consider the ball marked "1." The probability that Ann picks this ball is 1/10, and so the probability that Jane also picks this ball is 1/10. The probability that both women will pick the ball marked "1" is, therefore, 1/10 x 1/10 = 1/100.

Now consider the ball marked "2." Again, the probability that Ann picks this ball is 1/10, and the probability that Jane also picks this ball is 1/10. Thus, the probability that both women will pick the ball marked "2" is, 1/10 x 1/10 = 1/100.

We continue this for the balls marked 3, 4, 5, 6, 7, 8, 9, and 10. For each ball, the probability will be 1/100.

Since there are 10 balls, the probability that Ann and Jane will pick the same ball is, therefore, 10 x 1/100 = 10/100 = 1/10.

Answer: D

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
Legendary Member
Posts: 2568
Joined: Sun Oct 29, 2017 2:04 pmFollowed by: 6 members

by swerve » Mon Nov 12, 2018 9:45 am
For Ann or For Jane TOTAL cases = 10
For Ann favorable cases = 10
For Ann probability of picking the ball = 10/10 = 1
Now that she has replaced the ball TOTAL cases remains the same.
For Jane favorable case = 1 (If Ann had picked the ball number 2, then Jane has to pick only that ball, so only one choice for jane, hence only one fav case)

Probability for Jane picking the same ball = 1/10.
D is the right answer.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1