|
|
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.
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
75% (02:05) correct
25%
(02:25)
wrong
based on 1195
sessions
History
Date
Time
Result
Not Attempted Yet
Set S consists of all prime integers less than 10. If two numbers are chosen form set S at random, what is the probability that the product of these numbers will be greater than the product of the numbers which were not chosen?
(A) \(\frac{1}{3}\)
(B) \(\frac{2}{3}\)
(C) \(\frac{1}{2}\)
(D) \(\frac{7}{10}\)
(E) \(\frac{4}{5}\)
(A) \(\frac{1}{3}\)
(B) \(\frac{2}{3}\)
(C) \(\frac{1}{2}\)
(D) \(\frac{7}{10}\)
(E) \(\frac{4}{5}\)
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!
04 Feb 2012, 13:01
Kudos
Bookmarks
ashiima
Shortcut solution:
Set \(S\) is defined as \(\{2, 3, 5, 7\}\).
The most straightforward way to address this question is to recognize that we are selecting half of the numbers from the set (essentially dividing the group of four into two smaller groups of two). As a tie is not possible, the probability that the product of the numbers in one subgroup exceeds that in the other subgroup is \(\frac{1}{2}\), because neither subgroup is favored over the other in terms of probability.
Answer: C.
Hope it's clear.
15 Dec 2011, 22:57
Kudos
Bookmarks
The set will be
S = {2, 3, 5, 7}
Two numbers can be chosen from the set of four numbers in 4C2 = 6 ways.
Chosen Numbers ----- Product ----- Numbers not chosen ----- Product
2, 3 -------------------- 6 --------------- 5, 7 -------------------- 35
2, 5 -------------------- 10 --------------- 3, 7 -------------------- 21
2, 7 -------------------- 14 --------------- 3, 5 -------------------- 15
3, 5 -------------------- 15 --------------- 2, 7 -------------------- 14
3, 7 -------------------- 21 --------------- 2, 5 -------------------- 10
5, 7 -------------------- 35 --------------- 2, 3 -------------------- 6
Thus, from the table above, it is obvious that there are just three possible combinations in which the product of the numbers chosen is greater than the product of the numbers not chosen.
Thus the required probability = 3/6 = 1/2
Hope this is helpful.
S = {2, 3, 5, 7}
Two numbers can be chosen from the set of four numbers in 4C2 = 6 ways.
Chosen Numbers ----- Product ----- Numbers not chosen ----- Product
2, 3 -------------------- 6 --------------- 5, 7 -------------------- 35
2, 5 -------------------- 10 --------------- 3, 7 -------------------- 21
2, 7 -------------------- 14 --------------- 3, 5 -------------------- 15
3, 5 -------------------- 15 --------------- 2, 7 -------------------- 14
3, 7 -------------------- 21 --------------- 2, 5 -------------------- 10
5, 7 -------------------- 35 --------------- 2, 3 -------------------- 6
Thus, from the table above, it is obvious that there are just three possible combinations in which the product of the numbers chosen is greater than the product of the numbers not chosen.
Thus the required probability = 3/6 = 1/2
Hope this is helpful.
