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
GMATBootcamp Starts Sep 21
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE BOOTCAMP

Live Online Bootcamp Class with Top GMAT Expert Chris Peckover

Sep 21 to Oct 9, 2026

Schedule
Mon to Fri · 7:00 to 10:00 PM ET
Included
Live classes + 6 months of TTP OnDemand
  • Boost your GMAT score in less than one month in a live online class
  • 6 months access to TTP OnDemand video courses included
View bootcamp & 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.

Probabilty

Expert replies
by CSASHISHPANDAY » Wed Jul 25, 2012 9:28 am
n a certain game, a player begins with a bag containing tiles numbered 1 through 10, each of which has an equal probability of being selected. The player draws one tile. If the tile is even, the player stops. If not, the player draws another tile without replacing the first. If this second tile is even, the player stops. If not, the player draws a third tile-without replacing either of the first two tiles-and then stops. What is the probability that at the conclusion of the game, the sum of the tiles that the player has drawn is odd?

(A)5/18


(B) 13/36


(C)3/8


(D)23/36


(E)5/8
Join the discussion
Source: — Problem Solving |

by NicoleWhite » Wed Jul 25, 2012 1:19 pm
Hi. I think the answer is B.

I believe there are two ways to end up with a sum that is odd: draw an odd first, then an even, and stop, or draw all three odds. Any other way will end up with an even sum.

The probability of the first is: P(odd) * P(even) = 5/10 * 5/9 = 1/2 * 5/9 = 5/18
The probability of the second is: P(odd) * P(odd) * P(odd) = 5/10 * 4/9 * 3/8 = 12/144 = 1/12

Adding these two together: 5/18 + 1/12 = 78/216 = [spoiler]13/36[/spoiler]

I am new here so please let me know if I've done something incorrect. :D
Join the discussion