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 2212:30 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 1612:00 PM EDT
-11:59 PM EDT
You’re first to know: Save $200 on GMAT prep starting tomorrow, 4/13. Get 6 official practice tests and study with real questions. Be ready to save!
Apr 2301:30 AM EDT
-02:30 AM EDT
Struggling with GMAT Verbal as a non-native speaker? Harsh improved his score from 595 to 695 in just 45 days—and scored a 99 %ile in Verbal (V88)! Learn how smart strategy, clarity, and guided prep helped him gain 100 points.
Apr 2308:00 PM PDT
-09:00 PM PDT
Video explanations + diagnosis of 10 weakest areas + 150+ short videos + study plan!
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 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.
16 Sep 2004, 06:35
Kudos
Bookmarks
N
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct
0%
(00:00)
wrong
based on 0
sessions
History
Date
Time
Result
Not Attempted Yet
There are 10 ball in a bag,including 1 green ball and 1 yellow ball. Draw 3 balls from a bag without replacement, What are the possibilities that the 3 ball picked includes the green ball but not the yellow ball.
Is my approach to the problem correct? Read on...
- Prob of not selecting a yellow ball = 1 - (prob of selecting a yellow ball)
ie., 1- (1/10) = 9/10
- Prob of selecting a green ball = 1/10
- Prob of selecting 2 other balls from the pool = 8/9*7/8 (since it is without replacement)
so the answer is 9/10 * 1/10 * 8/9 * 7/8 = 7/100
Is my approach to the problem correct? Read on...
- Prob of not selecting a yellow ball = 1 - (prob of selecting a yellow ball)
ie., 1- (1/10) = 9/10
- Prob of selecting a green ball = 1/10
- Prob of selecting 2 other balls from the pool = 8/9*7/8 (since it is without replacement)
so the answer is 9/10 * 1/10 * 8/9 * 7/8 = 7/100
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 Problem Solving (PS) 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!
1. Total possibilities (or the denominator) is = 10C3 = 120.
2. Constraint is that we can pick the green and not the yellow. So we have 1C1*8C2 ways of doing this. Because there is only one greenball, it can be picked in 1C1 ways. We would have only 9 left now (having picked the green ball) - however, we should not pick yellow and hence we only have 8 balls left to choose the remaining 2 balls. So, 8C2 ways.
The result is 1C1*8C2/10C3 = 28/120 = 7/30.
2. Constraint is that we can pick the green and not the yellow. So we have 1C1*8C2 ways of doing this. Because there is only one greenball, it can be picked in 1C1 ways. We would have only 9 left now (having picked the green ball) - however, we should not pick yellow and hence we only have 8 balls left to choose the remaining 2 balls. So, 8C2 ways.
The result is 1C1*8C2/10C3 = 28/120 = 7/30.
