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.
Apr 2601:30 AM EDT
-02:30 AM EDT
Learn how Keshav, a Chartered Accountant, scored an impressive 705 on GMAT in just 30 days with GMATWhiz's expert guidance. In this video, he shares preparation tips and strategies that worked for him, including the mock, time management, and more.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2512:30 AM EDT
-01:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
May 0111:00 AM PDT
-12:00 PM PDT
Free- full-length test + 15 concept videos + 150+ short videos + study plan!
13 Apr 2022, 06:15
Kudos
Bookmarks
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
73% (02:01) correct
27%
(02:21)
wrong
based on 92
sessions
History
Date
Time
Result
Not Attempted Yet
Alan and Becky are among the ten students a teacher can choose from to form a six-student team. If all six students are chosen at random, what is the probability of choosing a team that includes Alan and Becky?
A. 1/9
B. 1/6
C. 2/9
D. 5/18
E. 1/3
Are You Up For the Challenge: 700 Level Questions
A. 1/9
B. 1/6
C. 2/9
D. 5/18
E. 1/3
Are You Up For the Challenge: 700 Level Questions
13 Apr 2022, 09:15
Kudos
Bookmarks
\(\frac{8}{10}+\frac{7}{9}+\frac{6}{8}+\frac{5}{7}+\frac{2}{6}+\frac{1}{5} = \frac{1}{45}\)
However, in doing this I have created a specific order and found the probability of choosing four students and then either Alan and Becky as the last two. To rectify this, I must multiply \(\frac{1}{45}\) by \(\frac{6!}{4!2!} = 15\)
\(\frac{1}{45} * 15 = \frac{15}{45} = \frac{1}{3}\)
Answer E.
However, in doing this I have created a specific order and found the probability of choosing four students and then either Alan and Becky as the last two. To rectify this, I must multiply \(\frac{1}{45}\) by \(\frac{6!}{4!2!} = 15\)
\(\frac{1}{45} * 15 = \frac{15}{45} = \frac{1}{3}\)
Answer E.
14 Apr 2022, 01:48
Kudos
Bookmarks
Solution=>
Total ways of choosing any team of 6-student team from 10 available students = 10C6 (order of choosing is not important)
Now we are asked to find the probability of choosing a team that includes Alan and Becky,
Lets suppose we have Alan and Becky in a team, now there are 4 positions remaining for that team and 8 students to choose from,
=> we can choose 4 students from a group of 8 in 8C4 ways
Therefore P(team with Alan and Becky) = Ways of choosing team with Alan and Becky/ Total ways of choosing the team
=> 8C4/10C6 = 1/3
Total ways of choosing any team of 6-student team from 10 available students = 10C6 (order of choosing is not important)
Now we are asked to find the probability of choosing a team that includes Alan and Becky,
Lets suppose we have Alan and Becky in a team, now there are 4 positions remaining for that team and 8 students to choose from,
=> we can choose 4 students from a group of 8 in 8C4 ways
Therefore P(team with Alan and Becky) = Ways of choosing team with Alan and Becky/ Total ways of choosing the team
=> 8C4/10C6 = 1/3
