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.
Jul 0911:00 AM PDT
-12:30 PM PDT
Welcome to the M7 MBA Spotlight Series, featuring MIT Sloan School of Management! This video is part of GMAT Club’s exclusive MBA Spotlight event, where top business schools meet prospective applicants like you.- Jun 30
08:00 AM PDT
-11:59 PM PDT
Join us June 30th for 2 weeks of GMAT questions (just 8 per day) to help with your prep and to compete with your fellow GMAT Clubbers for a chance to win a prize fund of $30,000 for you and your team-mates! - Jul 10
11:00 AM PDT
-11:00 AM PDT
In our conversation, Kishore talks about how strategically he used data analytics to identify high impact areas to focus on as well as his early morning study routine (3 AM prep!) that optimized his GMAT preparation. 
Jul 1211:00 AM IST
-01:00 PM IST
Attend this session to learn how to Deconstruct arguments, Pre-Think Author’s Assumptions, and apply Negation Test.
Jul 1311:00 AM EDT
-12:30 PM EDT
Looking for your GMAT motivation to break through the score plateau? Pragati improved her score by massive 160 points with strategic guidance and hard-work! Find out how personalized mentorship and a strong mindset can turn GMAT struggles into success.
Jul 1508:00 PM PDT
-09:00 PM PDT
Full-length FE mock with insightful analytics, weakness diagnosis, and video explanations!
25 Mar 2025, 00:30
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
61% (01:32) correct
39%
(01:42)
wrong
based on 172
sessions
History
Date
Time
Result
Not Attempted Yet
A jar contains 5 blue marbles, 4 red marbles and 3 white marbles. If three marbles are chosen from the jar at random without replacement, what is the probability that each one of the colors will be chosen?
A. 5/144
B. 1/22
C. 3/22
D. 3/11
E. 1/3
A. 5/144
B. 1/22
C. 3/22
D. 3/11
E. 1/3
25 Mar 2025, 00:53
Kudos
Bookmarks
Total marbles in the jar: \(5 + 4 + 3 = 12\)
We need to find the probability of selecting **one blue, one red, and one white** in any order.
Select 1 blue from 5
\(\frac{5}{12}\)
Select 1 red from 4 (now 11 marbles left)
\(\frac{4}{11}\)
Select 1 white from 3 (now 10 marbles left)
\(\frac{3}{10}\)
Probability for this specific sequence (Blue → Red → White):
\(\frac{5}{12} \times \frac{4}{11} \times \frac{3}{10} = \frac{60}{1320} = \frac{1}{22}\)
Since the selection can occur in any order, there are \(3! = 6\) ways to arrange the three colors.
Final probability:
\(6 \times \frac{1}{22} = \frac{6}{22} = \frac{3}{11}\)
Final Answer: \(\frac{3}{11}\)
We need to find the probability of selecting **one blue, one red, and one white** in any order.
Select 1 blue from 5
\(\frac{5}{12}\)
Select 1 red from 4 (now 11 marbles left)
\(\frac{4}{11}\)
Select 1 white from 3 (now 10 marbles left)
\(\frac{3}{10}\)
Probability for this specific sequence (Blue → Red → White):
\(\frac{5}{12} \times \frac{4}{11} \times \frac{3}{10} = \frac{60}{1320} = \frac{1}{22}\)
Since the selection can occur in any order, there are \(3! = 6\) ways to arrange the three colors.
Final probability:
\(6 \times \frac{1}{22} = \frac{6}{22} = \frac{3}{11}\)
Final Answer: \(\frac{3}{11}\)
25 Mar 2025, 01:09
Kudos
Bookmarks
We are looking for the probability that each one of colors is chosen without replacement.
We have 5 blue marbles, 4 red marbles, and 3 white marbles.
Choosing a blue, red, and white marble without replacement:
Since the probability of choosing any color combination is equal, it does not matter what scenario we begin with.
P(BRW) = 5/12 * 4/11 * 3/10 = 1/22
We must remember that BRW is just one way to select the color order. There will be a total 3! ways to select all possible color orders, so we multiple the probability of one color combination by 3!
1/22 * 3! = 1/22 * 6 = 3/11
Alternative solution:
We can use combinations to solve the probability as well.
P(each color) = # of favorable events / #number of total events
P(each color) = 5C1 * 4C1 * 3C1 / 12C3
5 * 4 * 3 / 12 * 11 * 10 / 3 * 2 * 1 =
5*4*3*6 / 12*11*10
= 3/11
Option D
We have 5 blue marbles, 4 red marbles, and 3 white marbles.
Choosing a blue, red, and white marble without replacement:
Since the probability of choosing any color combination is equal, it does not matter what scenario we begin with.
P(BRW) = 5/12 * 4/11 * 3/10 = 1/22
We must remember that BRW is just one way to select the color order. There will be a total 3! ways to select all possible color orders, so we multiple the probability of one color combination by 3!
1/22 * 3! = 1/22 * 6 = 3/11
Alternative solution:
We can use combinations to solve the probability as well.
P(each color) = # of favorable events / #number of total events
P(each color) = 5C1 * 4C1 * 3C1 / 12C3
5 * 4 * 3 / 12 * 11 * 10 / 3 * 2 * 1 =
5*4*3*6 / 12*11*10
= 3/11
Option D
