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

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

Probability & counting problem example

Expert replies
by soumya029 » Sun Sep 08, 2013 3:42 pm
Hi,

I was watching a video from Princeton review. It was about tackling hard maths problem. One of the probability problems explained on it has left me a little confused in relation to its answer.
the problem was "Slips of paper are numbered from 1000 to 2000.If one slip of paper is selected at random what is the probability of selecting a paper with exactly three identical digits."

The video mentioned 37/1001 as the correct answer.
I calculated a number of times and the answer came out to be 38/1001. Could anyone please explain which one of the two answers mentioned above is correct?

Thanks Soumya
Join the discussion
Source: — Problem Solving |

by Java_85 » Sun Sep 08, 2013 8:16 pm
1000 and 2000 2 numbers
1222 1333 1444 ... 8 possible numbers.

1011
1101
1110 We can put 23456789 instead of 0 too (But not 1) ==> 3*9=27 numbers

27+8+2=37

I also think it's 37/1001
Join the discussion

by [email protected] » Sun Sep 08, 2013 11:12 pm
Hi soumya029,

Go back and check your work. Did you include the number 1111 as an option? Because according to how the question is worded, you're NOT supposed to include that one.

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
Image
Join the discussion

by soumya029 » Mon Sep 09, 2013 10:34 am
Yeah you are right Rich, I did include 1111. Thanks for pointing it out.

Soumya
Join the discussion

by vipulgoyal » Mon Sep 09, 2013 10:07 pm
1000,1222,1333...1999 = 9
1x11 now as of now x could be 0,2,3...9
in each case the last 3 digits can be arranged in 3!/2! ways, 3 arrangments for each value of x =
3*9 = 27 arrangements
now if x = 1, one more arrangement 1111 is possible
hence no of arrangements = 9+27+1 = 37
Join the discussion