Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
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.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
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 Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemandStart 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

Wine bottles

Expert replies
by umaa » Thu Jul 17, 2008 1:37 am
I have 2 different wine bottles. If I want to drink one glass of wine every day, each bottle contains enough wine for 2 glasses, and I select one bottle of wine randomly each day, what is the probability that I finish one bottle completely before opening the second one?

a. 1/4
b. 5/6
c. 1/6
d. 1/3
e. 2/3

answer is, d
Join the discussion
Source: — Problem Solving |

by pepeprepa » Thu Jul 17, 2008 3:47 am
You have 2 bottles: A and B
You have 6 possible combinations of choosing A or B during the 4 days
and 2 good results (AABB and BBAA), you drink in the same one bottle the first two days

--> 2/6=1/3 -- >d
Join the discussion

by pepeprepa » Thu Jul 17, 2008 3:47 am
You have 2 bottles: A and B
You have 6 possible combinations of choosing A or B during the 4 days
and 2 good results (AABB and BBAA), you drink in the same one bottle the first two days

--> 2/6=1/3 -- >d
Join the discussion

by louvre » Thu Jul 17, 2008 4:55 am
I have tough time understanding this. Here is my logic!

There are 2 bottles (NOT glasses), the question is the probability of selecting the same bottle in first 2 days i.e day1 and day2.

So it's the probability of selecting a(any) bottle on the day 1 AND selecting the same bottle on the day 2.

P(a/any) = 1
p(the same next day) = 1/2.

Isn't it 1*1/2 = 1/2

If I look at in terms of glasses, I can pass through it, but only stuck when using the bottles.

Thanks for your clarifications.
Join the discussion

by pepeprepa » Thu Jul 17, 2008 6:55 am
I write all the possible combinations. You can draw a "tree" (don't know if that is the right term) to see all possiilities:
AABB
BBAA
ABAB
BABA
ABBA
BAAB

For your probabilistic logic I don't catch why we do not find 1/3, can't help you sorry.
Join the discussion

by pepeprepa » Thu Jul 17, 2008 6:56 am
I write all the possible combinations. You can draw a "tree" (don't know if that is the right term) to see all possiilities:
AABB
BBAA
ABAB
BABA
ABBA
BAAB

For your probabilistic logic I don't catch why we do not find 1/3, can't help you sorry.
Join the discussion

by anksanks » Thu Jul 17, 2008 7:50 am
let the bottle be labelled a and b
and glasses be labeled a1, a2 for bottle a and b1 and b2 for bottle b

on first day,
probability for a1 = 1/4
probability for a2 = 1/4
adding up for a = 1/4 + 1/4 = 1/2

on day two,
assuming a1 was selected on day 1,
probability for a2 = 1/3
assuming a2 was selected on day 1,
probability for a1 = 1/3
total probability for a = 1/3 + 1/3 = 2/3

for two days, probability for a will be
1/2 * 2/3 = 1/3

Please correct me if I am wrong...
Shape me with your bounds,
Wrap me all around,
Define this adorable thought,
And break the silence.
(c) Anks
Join the discussion

by pepeprepa » Thu Jul 17, 2008 8:34 am
anksanks, what i don't catch is why the probability of getting a2, assuming you got a1 the first day, is 1/3
I mean you do not choose a2 among 3 glasses of wine (given a1 is finished) but you choose between a bottle and another one, so 2 choices.
Tell me if I am wrong, I had a long day
Join the discussion

by anksanks » Thu Jul 17, 2008 8:51 am
What i understood from the problem is it is taking individual glasses and not bottles into consideration.
like there are two red balls and two green balls and not red color or green color.

The answers will totaly change if we assume it that way.

My first understanding was to divide it into four individual glasses like we divide it in case of colored balls or other probability issues.

I hope it helps.
Shape me with your bounds,
Wrap me all around,
Define this adorable thought,
And break the silence.
(c) Anks
Join the discussion