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 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 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.
03 Jan 2021, 11:24
Kudos
Bookmarks
B
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% (01:23) correct
39%
(01:17)
wrong
based on 115
sessions
History
Date
Time
Result
Not Attempted Yet
From the even numbers between 1 and 9, two different even numbers are to be chosen at random, without replacement. What is the probability that their sum will be 8?
A. 1/12
B. 1/6
C. 1/4
D. 1/3
E. 1/2
A. 1/12
B. 1/6
C. 1/4
D. 1/3
E. 1/2
04 Jan 2021, 03:06
Kudos
Bookmarks
Even Number between \(1\) and \(9\) are \(2,4,6,8\).
The sum of the two numbers should be \(8\), and the possible ways are \((2,6)\), or, \((6,2)\)
Probability of selecting the \(1_{st}\) number (between \(2\) or \(6\)) \(=\frac{2}{4}=\frac{1}{2}\)
Probability of selecting the \(2_{nd}\) number (between \(4,6,8\) or \(2,4,8\)) \(=\frac{1}{3}\)
The required probability \(= \frac{1}{2}∗\frac{1}{3}=\frac{1}{6}\)
IMO Ans B
The sum of the two numbers should be \(8\), and the possible ways are \((2,6)\), or, \((6,2)\)
Probability of selecting the \(1_{st}\) number (between \(2\) or \(6\)) \(=\frac{2}{4}=\frac{1}{2}\)
Probability of selecting the \(2_{nd}\) number (between \(4,6,8\) or \(2,4,8\)) \(=\frac{1}{3}\)
The required probability \(= \frac{1}{2}∗\frac{1}{3}=\frac{1}{6}\)
IMO Ans B
10 Feb 2021, 00:22
Kudos
Bookmarks
Bunuel
Ways to choose two even numbers out of four(2,4,6,8) = 1st*2nd = 4*3 = 12
Even numbers that sum 8 are 2 and 6.
(4 is not possible for once 4 is choose first the other has to be 4 itself the 2nd time but it would required a replacement of 4, which is not possible AND 8 is also not possible since 0 is not available).
Ways to choose two even numbers that sum 8 = 2*1 = 2
Probability = \(\frac{2}{12} = \frac{1}{6}\)
Answer B.
