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 Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 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
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
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 Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget 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 - choosing colours/cards

Expert replies
by StoneBlack » Tue Sep 06, 2011 8:28 pm
1. Kurt, a painter, has 9 jars of paint:
4 are yellow
2 are red
rest are brown
Kurt will combine 3 jars of paint into a new container to make a new color, which he will name accordingly to the following conditions:

Brun Y if the paint contains 2 jars of brown paint and no yellow
Brun X if the paint contains 3 jars of brown paint
Jaune X if the paint contains at least 2 jars of yellow
Jaune Y if the paint contains exactly 1 jar of yellow

What is the probability that the new color will be Jaune

a) 5/42
b) 37/42
c) 1/21
d) 4/9
e) 5/9

I have a question about how to choose the colours after some answers. i would like to contrast this question with another one:

2. In how many ways can one choose 6 cards from a normal deck of cards so as to have all suits present?
a. (13^4) x 48 x 47
b. (13^4) x 27 x 47
c. 48C6
d. 13^4
e. (13^4) x 48C6

Again this involves how you choose the cards.
Join the discussion
Source: — Problem Solving |

by Anurag@Gurome » Tue Sep 06, 2011 8:43 pm
Kurt, a painter, has 9 jars of paint:
4 are yellow
2 are red
rest are brown
Kurt will combine 3 jars of paint into a new container to make a new color, which he will name accordingly to the following conditions:

Brun Y if the paint contains 2 jars of brown paint and no yellow
Brun X if the paint contains 3 jars of brown paint
Jaune X if the paint contains at least 2 jars of yellow
Jaune Y if the paint contains exactly 1 jar of yellow

What is the probability that the new color will be Jaune

a) 5/42
b) 37/42
c) 1/21
d) 4/9
e) 5/9

The probability that the new color will be Jaune is possible in the following situations:
(1) There are at least 2 jars of yellow paint
(2) There is exactly 1 jar of yellow paint

Now let's first find the probability that there are no jars of yellow paint = 5/9 * 4/8 * 3/7 = 5/42

Therefore, probability that some jars are of yellow paint = 1 - 5/42 = 37/42

The correct answer is B.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Join the discussion

by StoneBlack » Tue Sep 06, 2011 9:03 pm
Thanks Anurag for your reply.

Suppose we do this by choosing colours.

1. 2 yellow means..

a> there can be all three Y => 4c3
OR
b> 2 Y and 1 out of 2 R and 3 B => 4c2 x 5c1

Total 34

2.This has exactly 1 Y and remaining 2 out of 5 = > 4c1 x 5c2

Total 40

Likely possibilities : 34 + 40 = 74
Total possibilities = (9!/3!6!) = 84
Hence probability = 74/84 = 37/42

My question is when choosing for case 1 (Jaune X)
can we not choose in this way : 2 colours of Yellow from 4, 4C2
And Rest one colour can be from the remaining 7 colours left hence, 7C1
or 4C2 * 7C1. Where is the mistake in this?
Join the discussion

by saketk » Wed Sep 07, 2011 8:29 am
StoneBlack wrote:Thanks Anurag for your reply.

Suppose we do this by choosing colours.

1. 2 yellow means..

a> there can be all three Y => 4c3
OR
b> 2 Y and 1 out of 2 R and 3 B => 4c2 x 5c1

Total 34

2.This has exactly 1 Y and remaining 2 out of 5 = > 4c1 x 5c2

Total 40

Likely possibilities : 34 + 40 = 74
Total possibilities = (9!/3!6!) = 84
Hence probability = 74/84 = 37/42

My question is when choosing for case 1 (Jaune X)
can we not choose in this way : 2 colours of Yellow from 4, 4C2
And Rest one colour can be from the remaining 7 colours left hence, 7C1
or 4C2 * 7C1. Where is the mistake in this?
No, you cannot do that.. you cannot have 7 jars for next selection. Because your case 1 says 2Y + 1 jar of any other color..

So, if you choose 7 then it will also include the 2 Yellow jars. Which is incorrect.

Case 1:- 2Y jars+1 of any other color
case 2:- all 3 are yellow..

I hope this helps.
Join the discussion

by saketk » Wed Sep 07, 2011 8:59 am
Answer to your second question--
2. In how many ways can one choose 6 cards from a normal deck of cards so as to have all suits present?
a. (13^4) x 48 x 47
b. (13^4) x 27 x 47
c. 48C6
d. 13^4
e. (13^4) x 48C6

Again this involves how you choose the cards.

Total cards of each type 13

Spade - 13
Diamond - 13
Hearts - 13
Clubs - 13

Clearly, in order to have all types in your selection, you must have 1 of each type first..

So, 1 of each type can be selected in 13 ways each

Rest 2 can be selected out of remaining 48 cards in 48*47 ways..

Therefore, the total comes out to be ..

(13)^4*48*47

Correct Answer -- A.
Join the discussion

by navami » Thu Sep 08, 2011 5:17 am
Second question : 13^4 *48*47 ---> choice A
This time no looking back!!!
Navami
Join the discussion