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.
Nov 2007:30 AM PST
-08:30 AM PST
Learn what truly sets the UC Riverside MBA apart and how it helps in your professional growth
Nov 2001:30 PM EST
-02:30 PM IST
Learn how Kamakshi achieved a GMAT 675 with an impressive 96th %ile in Data Insights. Discover the unique methods and exam strategies that helped her excel in DI along with other sections for a balanced and high score.
Nov 2206:30 AM PST
-08:30 AM PST
Let’s dive deep into advanced CR to ace GMAT Focus! Join this webinar to unlock the secrets to conquering Boldface and Paradox questions with expert insights and strategies. Elevate your skills and boost your GMAT Verbal Score now!
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2407:00 PM PST
-08:00 PM PST
Full-length FE mock with insightful analytics, weakness diagnosis, and video explanations!
Nov 2510:00 AM EST
-11:00 AM EST
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.
bipolarbear
Joined: 11 Dec 2008
Last visit: 16 Sep 2013
Posts: 353
Given Kudos: 12
Location: United States
Schools: HBS '15 (D) Stanford '15 (WL) Wharton '15 (WL)
GMAT 1: 760 Q49 V44
GPA: 3.9
06 Aug 2009, 10:38
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
51% (01:52) correct
49%
(01:52)
wrong
based on 262
sessions
History
Date
Time
Result
Not Attempted Yet
If two integers are selected, one after the other with replacement, from a set of integers from 20 to 29, inclusive, what is the probability that one of the selected integers is a prime number, and the other is a multiple of 3?
A. 0.06
B. 0.12
C. 0.15
D. 0.18
E. 0.20
A. 0.06
B. 0.12
C. 0.15
D. 0.18
E. 0.20
Experience GMAT Club Test Questions
Yes, you've landed on a GMAT Club Tests question
Craving more? Unlock our full suite of GMAT Club Tests here
Want to experience more? Get a taste of our tests with our free trial today
Rise to the challenge with GMAT Club Tests. Happy practicing!
20 Sep 2023, 08:45
Kudos
Bookmarks
If two integers are selected, one after the other with replacement, from a set of integers from 20 to 29, inclusive, what is the probability that one of the selected integers is a prime number, and the other is a multiple of 3?
A. 0.06
B. 0.12
C. 0.15
D. 0.18
E. 0.20
Start by identifying the prime integers and multiples of 3 in the range 20 to 29. The prime integers are 23 and 29, while the multiples of 3 are 21, 24, and 27.
We can then consider two cases:
Case 1: The first number selected is prime, and the second is a multiple of 3. The probability of this is (2/10) * (3/10) = 0.06. Notice, that since we are selecting two integers with replacement, the denominator for selection case is 10 (the total number of integers in the given range).
Case 2: The first number selected is a multiple of 3, and the second is prime. The probability of this is (3/10) * (2/10) = 0.06.
Therefore, the total probability is the sum of the probabilities of the above two cases: 0.06 + 0.06 = 0.12.
Answer: B
A. 0.06
B. 0.12
C. 0.15
D. 0.18
E. 0.20
Start by identifying the prime integers and multiples of 3 in the range 20 to 29. The prime integers are 23 and 29, while the multiples of 3 are 21, 24, and 27.
We can then consider two cases:
Case 1: The first number selected is prime, and the second is a multiple of 3. The probability of this is (2/10) * (3/10) = 0.06. Notice, that since we are selecting two integers with replacement, the denominator for selection case is 10 (the total number of integers in the given range).
Case 2: The first number selected is a multiple of 3, and the second is prime. The probability of this is (3/10) * (2/10) = 0.06.
Therefore, the total probability is the sum of the probabilities of the above two cases: 0.06 + 0.06 = 0.12.
Answer: B
General Discussion
14 Jan 2016, 05:20
Kudos
Bookmarks
shreyashid
No. Note that the numbers are picked with replacement. 10C2 gives you the number of ways of picking the two numbers one after the other without replacement.
You can pick the prime number in 2 ways and then the multiple of 3 in 3 ways. This gives you 2*3 = 6 ways.
But you can also pick the multiple of 3 first and then the prime number. This gives you another 3*2 = 6 ways.
No of ways of picking the first number is 10 and the second number is 10 again. So total ways = 100 ways
Probability = 12/100
or use probability
20-29 are 10 numbers of which 2 are prime (23 and 29) and 3 are multiples of 3 (21, 24 and 27)
First, you can select a prime with probability 2/10 and then a multiple of 3 with probability 3/10.
Or you can select a multiple of 3 with probability 3/10 and then a prime with probability 2/10.
Total probability = (2/10)*(3/10) + (3/10)*(2/10) = 3/25
