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.
12 May 2020, 07:56
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
50% (02:22) correct
50%
(02:09)
wrong
based on 146
sessions
History
Date
Time
Result
Not Attempted Yet
A die is thrown three times and the sum of the three numbers is found to be 15. The probability that the first throw was a four is
A. 1/10
B. 1/6
C. 1/5
D. 1/4
E. 1/3
Are You Up For the Challenge: 700 Level Questions
A. 1/10
B. 1/6
C. 1/5
D. 1/4
E. 1/3
Are You Up For the Challenge: 700 Level Questions
12 May 2020, 08:11
Kudos
Bookmarks
a+b+c=15
\(\frac{a+b+c}{3} = 5\)
The largest among them can be 5 or 6
Case 1- When largest among them is 5.
Possible solution = (5,5,5)
Total arrangement\(= \frac{3!}{3!} = 1\)
Case 2- When largest among them is 6.
Possible solutions = (4,5,6) and (3,6,6)
Total arrangements \(= 3!+\frac{3!}{2!} = 9\)
Total cases = 1+9=10
Favorable cases = 2 [(4,5,6) and (4,6,5)]
Probability = 2/10 = 1/5
\(\frac{a+b+c}{3} = 5\)
The largest among them can be 5 or 6
Case 1- When largest among them is 5.
Possible solution = (5,5,5)
Total arrangement\(= \frac{3!}{3!} = 1\)
Case 2- When largest among them is 6.
Possible solutions = (4,5,6) and (3,6,6)
Total arrangements \(= 3!+\frac{3!}{2!} = 9\)
Total cases = 1+9=10
Favorable cases = 2 [(4,5,6) and (4,6,5)]
Probability = 2/10 = 1/5
Bunuel
12 May 2020, 08:19
Kudos
Bookmarks
Bunuel
15 can be formed in following ways
{5, 5, 5} - Total Orders of 5, 5, 5 = 1
{3, 6, 6} - Total Orders of 3, 6, 6= 3!/2! = 3
{4, 6, 5} - Total Orders of 4, 6, 5= 3! = 6
Total Outcomes = 1+3+6 = 10
Favorable outcomes = {4, 5, 6} or {4, 6, 5} = 2
Required probability = 2/10 = 1/5
Answer: Option C
