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.

Probability

Expert replies
by ajmoney09 » Tue Jun 09, 2009 8:19 pm
A roulette wheel consists of 38 slots. 36 of these are
numbered 1 through 36 and colored red or black so
that there are nine red even-numbered slots, nine black
odd-numbered slots, etc. These slots occur with equal
probability. The two slots marked 0 or 00 are each
three times as likely to occur as any one of the other
36. What is the probability that a red even number, a
red 23, or a 00 will occur on one roll of the wheel?


The answer is 13/42.

Someone please explain this to me, thanks.
Join the discussion
Source: — Problem Solving |

by tohellandback » Tue Jun 09, 2009 10:51 pm
How can there be a red numbered 23? it must be even. or don't I understand the question?

if it were just 23 and not red 23:

total probable events are 36+(2*3)
because 0 and 00 are three times as likely to occur as anyof the other 36

probabily of a red= 9/42

probability of a 23=1/42

probabilty of 00=3/42

Adding up-13/42

adding up- 13/42
The powers of two are bloody impolite!!
Join the discussion