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.
May 0510:00 AM PDT
-11:00 AM PDT
GMAT Inequalities is a high-frequency topic in GMAT Quant, but many students struggle because the concepts behave differently from standard algebra. Understanding the right rules, patterns, and edge cases can significantly improve both speed and accuracy.- May 06
08:30 AM PDT
-09:30 AM PDT
In Episode 3 of our GMAT Ninja Critical Reasoning series, we tackle Discrepancy, Paradox, and Explain an Oddity questions. You know the feeling: the passage gives you two facts that seem completely contradictory.... - May 08
08:00 AM PDT
-08:30 AM PDT
Join the special YouTube live-stream for selecting the winners of GMAT Club MBA Scholarships sponsored by Juno live. Watch who gets these coveted MBA scholarships offered by GMAT Club and Juno.
27 Aug 2009, 14:50
Kudos
Bookmarks
Question: There are 6 team members. 4 have to be selected. What is probability that 2 particular members, say X and Y, are both selected.
Approach 1:
Total possible outcomes = 6! / (4! x 2!) = 15 teams
Winning outcomes where 2 particular members are on team= 4! /(2! x 2!) = 6
Therefore probability that 2 particular members on team = 6/15 = 2/5
Approach 2:
let X be selected first, Y second, "Some other" third, "Some other" fourth - to make a four member team out of 6.
Therefore probability = 1/6 x 1/5 x 4/4 x 3/3 = 1/30
But since order does not matter, actual probability = 1/30 x 4! = 4/5
OA : 2/5
Whats wrong with approach 2
Approach 1:
Total possible outcomes = 6! / (4! x 2!) = 15 teams
Winning outcomes where 2 particular members are on team= 4! /(2! x 2!) = 6
Therefore probability that 2 particular members on team = 6/15 = 2/5
Approach 2:
let X be selected first, Y second, "Some other" third, "Some other" fourth - to make a four member team out of 6.
Therefore probability = 1/6 x 1/5 x 4/4 x 3/3 = 1/30
But since order does not matter, actual probability = 1/30 x 4! = 4/5
OA : 2/5
Whats wrong with approach 2
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
27 Aug 2009, 16:00
Kudos
Bookmarks
Hi,
I'm not familiarized with this method (2) but I think you just have to do the following:
\(\frac{1}{30} * \frac{4!}{2!} = \frac{12}{30} = \frac{2}{5}\)
Good luck tomorrow!
I'm not familiarized with this method (2) but I think you just have to do the following:
\(\frac{1}{30} * \frac{4!}{2!} = \frac{12}{30} = \frac{2}{5}\)
Good luck tomorrow!
