Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Aug 2710:00 AM EDT
-11:00 AM EDT
We’ve worked incredibly hard to build TTP into the best test prep experience possible, and it would mean a lot to us to win Newsweek’s 2026 Readers’ Choice Award for Best Test Prep. If TTP has helped you, we’d be incredibly grateful for your vote.
Sep 0410:00 AM PDT
-11:00 AM PDT
What does test anxiety really look like on the GMAT? For many test takers, it does not look like a panic attack. Instead, it shows up as brain fog, rereading, negative comparisons, mental noise....
Sep 0609:30 AM EDT
-12:30 PM EST
Top scores are possible when you enroll in a powerful EA course, taught live online + 6 months access to TTP OnDemand video courses included! Perfect class schedule and easy course access for working professionals. Class starts Sundays Sept. 6, 2026.- Sep 07
12:01 AM PDT
-11:59 PM PDT
On the 7th, GMAT Club Tests will be Free! Including all the Quizzes, Questions, and Tests. 12 AM - 11:59 PM PDT - Sep 09
07:00 AM PDT
-10:00 AM PDT
Meet AdComs and explore top Master’s programs - MiM, MiF, MSc, MSBA and more. - Application Fee Waivers - Free 1-Week of GMAT Club Tests: - Master's Application Toolkit - Grand Prize Giveaway 
Sep 1507:00 PM EDT
-09:00 PM EST
Elite scores are possible when you enroll in a powerful GMAT course, taught live online + 6 months access to TTP OnDemand video courses included! Class starts Tues/Thurs Sept. 15, 2026 - Nov. 15, 2027, 7:00pm-9:00pm EST
Sep 2108:00 PM EDT
-10:00 PM EDT
Boost your GMAT score in less than one month in a live online class + 6 months access to TTP OnDemand video courses included! Class starts Mon, Tues, Wed, Thur, Fri Sept. 21, 2026 - Oct. 9, 2027, 7:00pm-10:00pm EST
14 Jul 2026, 09:37
Kudos
Bookmarks
R - 5, B - 3, G - 2
4 rounds
needed: 4 choices are NOT exactly 2 from R and exactly 2 from B
So i will find 1 - 'exactly 2 from R and 2 from B to get the answer'
select 2 rounds 4C2 and select only members from Team red for these 2 rounds. Since each member can be selected more than once , you get 4C2 * 5 * 5
And then other 2 choices from Blue team will be 3*3
So overall prob for 4 choices from Blue and Red is 4C2*5*5*3*3/10*10*10*10 = 27/200
Now 1 - 27/200 = 173/200 D)
4 rounds
needed: 4 choices are NOT exactly 2 from R and exactly 2 from B
So i will find 1 - 'exactly 2 from R and 2 from B to get the answer'
select 2 rounds 4C2 and select only members from Team red for these 2 rounds. Since each member can be selected more than once , you get 4C2 * 5 * 5
And then other 2 choices from Blue team will be 3*3
So overall prob for 4 choices from Blue and Red is 4C2*5*5*3*3/10*10*10*10 = 27/200
Now 1 - 27/200 = 173/200 D)
14 Jul 2026, 09:43
Kudos
Bookmarks
The no of choices for exactly 2 from Team Red and 2 from Team Blue is= 5*5*3*3 (repetition is allowed)
Sample space= 10*10*10*10
.: Probability of that configuration is=5*5*3*3/(10*10*10*10)=9/400
.: Probability of not having that configuration is, 1-9/400 =391/400
Sample space= 10*10*10*10
.: Probability of that configuration is=5*5*3*3/(10*10*10*10)=9/400
.: Probability of not having that configuration is, 1-9/400 =391/400
Bunuel
14 Jul 2026, 10:03
Kudos
Bookmarks
Here we have 10 Contestants divided into 3 Teams:
Red -> 5
Blue -> 3
Green ->2
The desired probability is what is the probability that 4 choices of the selected rounds are not made up exactly of 2 from Team Red and 2 from Team Blue.
We can re-phrase the Probability as 1 - [2 Nos. of Candidates selected exactly from Team Red and Team Blue]. Remember doing this at the end else you'll be caught by the trick in the question.
Now coming to the candidates selected exactly from Team Red and Team Blue:
Total Outcomes Over all 4 Rounds = 10C1 X 10C1 X 10C1 X 10C1 = 10,000
Generally Probability = Favourable Outcomes / Total Outcomes
In this case it is, 1 - [Favourable / Total Outcomes], because we need to find the probability of NOT being from the selection.
Anyway, Favourable Outcomes is:
No. of Rounds in which the two teams are required:
4 Rounds; Each Round needs Red / Blue so something like RRBB or RBRB or BRBR or BBRR..... and so on.
The total number of these combinations are 4C2.
And the selection in each Round can be done in 5C1 X 5C1 X 3C1 X 3C1 no matter in which order they are ultimately arranged in each Round. If 2 are Red, the other 2 are automatically blue.
Hence the favourable outcomes are 4C2 X 5C1 X 5C1 X 3C1 X 3C1 = 1,350 Outcomes
Hence Probability = 1 - [1,350 / 10,000] = 1 - [27/200] = 173/200 => Option D is the answer.
Red -> 5
Blue -> 3
Green ->2
The desired probability is what is the probability that 4 choices of the selected rounds are not made up exactly of 2 from Team Red and 2 from Team Blue.
We can re-phrase the Probability as 1 - [2 Nos. of Candidates selected exactly from Team Red and Team Blue]. Remember doing this at the end else you'll be caught by the trick in the question.
Now coming to the candidates selected exactly from Team Red and Team Blue:
Total Outcomes Over all 4 Rounds = 10C1 X 10C1 X 10C1 X 10C1 = 10,000
Generally Probability = Favourable Outcomes / Total Outcomes
In this case it is, 1 - [Favourable / Total Outcomes], because we need to find the probability of NOT being from the selection.
Anyway, Favourable Outcomes is:
No. of Rounds in which the two teams are required:
4 Rounds; Each Round needs Red / Blue so something like RRBB or RBRB or BRBR or BBRR..... and so on.
The total number of these combinations are 4C2.
And the selection in each Round can be done in 5C1 X 5C1 X 3C1 X 3C1 no matter in which order they are ultimately arranged in each Round. If 2 are Red, the other 2 are automatically blue.
Hence the favourable outcomes are 4C2 X 5C1 X 5C1 X 3C1 X 3C1 = 1,350 Outcomes
Hence Probability = 1 - [1,350 / 10,000] = 1 - [27/200] = 173/200 => Option D is the answer.
