|
Author |
Message |
|
VP
Joined: 18 May 2008
Posts: 1305
Followers: 9
Kudos [?]:
57
[0], given: 0
|
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} Source: GMAT Club Tests - hardest GMAT questions What is wrong with my approach? The set consists of (2,3,5,7) Niow acc to soncition the nos satisfying are (5,7), (3,5) and (3,7) Thus the probabiliy is (1/4*1/4)+(1/4*1/4)+(1/4*1/4)=3/16 But this is not the ans 
|
|
|
|
|
|
|
|
|
SVP
Joined: 17 Jun 2008
Posts: 1593
Followers: 7
Kudos [?]:
131
[1] , given: 0
|
1
This post received
KUDOS
It should be (1/4*1/3)+(1/4*1/3)+(1/4*1/3) = 3/12 = 1/4.
But, the sequence is immaterial here as we are interested in sum. Hence, (5,3) and (3,5) will give the same result. Thus, multiply 1/4 by 2 and 1/2 should be the answer.
Alternatively, I will solve it as follows:
favorable outcome = 3 (as you have listed earlier).
Since, order does not matter, total outcome = 4C2 = 6
Hence, probability = 3/6 = 1/2
|
|
|
|
|
|
Manager
Joined: 14 Oct 2008
Posts: 160
Followers: 1
Kudos [?]:
17
[0], given: 0
|
I would go with exactly the same approach as the second one stated by scthakur.
Total outcomes = 2 numbers out of 4 hence 4C2 =6
Possible outcomes = 2 numbers out of 3 hence 3C2 = 3
Probability = 3/6 = 1/2
Whats the QA ?
|
|
|
|
|
|
Manager
Joined: 18 Nov 2008
Posts: 118
Followers: 1
Kudos [?]:
8
[0], given: 0
|
scthakur, nice explanation
C
|
|
|
|
|
|
Intern
Joined: 25 Nov 2008
Posts: 7
Followers: 0
Kudos [?]:
6
[0], given: 0
|
We now the total possibilities is 4c2
List out the possibilities
Chossen Not choosen
23 57 - no
25 37 -no
27 35 - no
35 27 -yes
37 25 - yes
57 23 - yes
4c2=6
3/6 = 1/2
|
|
|
|
|
|
Manager
Joined: 29 Nov 2009
Posts: 109
Location: United States
Followers: 1
Kudos [?]:
21
[0], given: 5
|
The total number of combinations is the following: 4!/(2!*2!) = 6
The only ones that work are these: 3*5, 3*7, 5*7
3/6 = 1/2
|
|
|
|
|
|
Current Student
Affiliations: CFA
Joined: 21 Dec 2008
Posts: 384
Location: United States (NY)
Schools: Columbia - Class of 2013
GMAT 1: 710 Q45 V43
GMAT 2: 760 Q49 V45
Followers: 21
Kudos [?]:
96
[1] , given: 61
|
1
This post received
KUDOS
All solutions that do not include a 2 satisfy the condition. So basically we just need to figure out the odds of not drawing a 2. 3/4 * 2/3 = 6/12 = 1/2 So C.
_________________
Find out what's new at GMAT Club - latest features and updates
|
|
|
|
|
|
Manager
Joined: 04 Dec 2009
Posts: 75
Location: INDIA
Followers: 2
Kudos [?]:
6
[0], given: 4
|
Ans : C there are only two possibal out come and have 50% chance of geting either one
_________________
MBA (Mind , Body and Attitude )
|
|
|
|
|
|
Manager
Joined: 26 Dec 2009
Posts: 151
Location: United Kingdom
Concentration: Strategy, Technology
GMAT 1: 500 Q45 V16
WE: Consulting (Computer Software)
Followers: 2
Kudos [?]:
12
[0], given: 10
|
C for me as well. 
|
|
|
|
|
|
Intern
Joined: 16 Dec 2010
Posts: 7
Followers: 0
Kudos [?]:
1
[0], given: 1
|
i think the ans is 1/3
Pls check whether my understanding is correct
We need to select two numbers from 1, 2, 3 ,5, 7 ( left out are 4, 6, 8, 9)
There are only two possibility, either its (5,7) or (7,5), for which product is 35 and is greater than the product of (4,6) / (6,4) rest all are lower than any combination from left out series.
So p= 2/4C2 = 2/6 = 1/3 Ans
|
|
|
|
|
|
Manager
Joined: 01 Nov 2010
Posts: 204
Location: India
Concentration: Technology, Marketing
GMAT Date: 08-27-2012
GPA: 3.8
WE: Marketing (Manufacturing)
Followers: 5
Kudos [?]:
10
[0], given: 26
|
GUD eXPLANATION.. its C.
_________________
kudos me if you like my post.
Attitude determine everything.
all the best and God bless you.
|
|
|
|
|
|
Manager
Joined: 24 Jul 2009
Posts: 197
Location: Anchorage, AK
Schools: Mellon, USC, MIT, UCLA, NSCU
Followers: 4
Kudos [?]:
17
[0], given: 10
|
C. The limitations of the set allow you to list the specific product options and compare them. I'm someone who's skilling up in quant and since I know that I make simple mistakes, I opted to actually write out the product options.
_________________
Reward wisdom with kudos. 
|
|
|
|
|
|
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11628
Followers: 1802
Kudos [?]:
9618
[1] , given: 829
|
1
This post received
KUDOS
ritula wrote: 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} Source: GMAT Club Tests - hardest GMAT questions S={2, 3, 5, 7} The simplest way would be to realize that we choose half of the numbers ( basically we divide the group of 4 into two smaller groups of 2) and since a tie is not possible then the probability that the product of the numbers in either of subgroup is more than that of in another is 1/2 (the probability doesn't favor any of two subgroups). Answer: C.
_________________
PLEASE READ AND FOLLOW: 11 Rules for Posting!!!
RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory
COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!
DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!
What are GMAT Club Tests?
25 extra-hard Quant Tests
Find out what's new at GMAT Club - latest features and updates
|
|
|
|
|
|
Kaplan GMAT Instructor
Joined: 21 Jun 2010
Posts: 148
Location: Toronto
Followers: 31
Kudos [?]:
89
[0], given: 0
|
zahuruddin wrote: i think the ans is 1/3
Pls check whether my understanding is correct
We need to select two numbers from 1, 2, 3 ,5, 7 ( left out are 4, 6, 8, 9)
There are only two possibility, either its (5,7) or (7,5), for which product is 35 and is greater than the product of (4,6) / (6,4) rest all are lower than any combination from left out series.
So p= 2/4C2 = 2/6 = 1/3 Ans Hi! 1 isn't a prime number, which is the first reason that you're deriving a different answer. The second reason is that you've misinterpreted the question - we're not comparing the product of the two numbers we select to the product of the non-primes less than 10; we're comparing the product of the two numbers we select to the product of the other primes less than 10. So, our set is {2, 3, 5, 7} and we're comparing the products of two pairs of numbers from within the set. The quickest ways to solve are to either use Bunuel's intuitive approach (we're comparing 2 vs 2 with no ties, so symmetry will lead to half the pairs being greater than the other half) or via brute force. Since there are only 6 possibilities, it doesn't take long to count them out: 2*3 vs 5*7 - not greater 2*5 vs 3*7 - not greater 2*7 vs 3*5 - not greater 3*5 vs 2*7 - greater 3*7 vs 2*5 - greater 5*7 vs 3*5 - greater Probability = (# of desired outcomes)/(total # of possibilities) = 3/6 = 1/2
_________________
Stuart Kovinsky
stuart.kovinsky@kaplan.com
Kaplan Test Prep & Admissions
Toronto Office
1-800-KAP-TEST
http://www.kaptest.com/GMAT
Prepare with Kaplan and save $150 on a course!
Kaplan Reviews
|
|
|
|
|
|
Intern
Joined: 28 Dec 2011
Posts: 31
GMAT 1: 750 Q50 V41
Followers: 0
Kudos [?]:
9
[0], given: 21
|
Counting 1 as a prime number is exactly the mistake I made... Funnily enough, they have an answer choice that corresponds to this particular way of "solving" the question.
|
|
|
|
|
|
Manager
Joined: 15 Dec 2011
Posts: 174
GMAT 1: 730 Q50 V39
GMAT 2: Q V
GPA: 3.9
Followers: 1
Kudos [?]:
27
[0], given: 6
|
C
total ways of choosing 2 out of 4 = !4/(!2*!2) i.e. 6
total ways the product can be greater than the other two = (3,5), (5,7), (7,3) = 3
answer=3/6 = 1/2
|
|
|
|
|
|
Manager
Joined: 08 Sep 2011
Posts: 82
Concentration: Finance, Strategy
Followers: 3
Kudos [?]:
0
[0], given: 5
|
I solved by calculating how many solutions do not involve 2.
1 - 1/4 * 1 - 1/3 = 6 /12 = 1/2. answer is C
|
|
|
|
|
|
Manager
Joined: 21 Nov 2010
Posts: 141
Followers: 0
Kudos [?]:
2
[0], given: 12
|
C. The Kaplan Instructor explained it pretty well.
|
|
|
|
|
|
Intern
Joined: 19 Dec 2009
Posts: 37
Followers: 0
Kudos [?]:
9
[0], given: 8
|
Moss wrote: All solutions that do not include a 2 satisfy the condition. So basically we just need to figure out the odds of not drawing a 2.
3/4 * 2/3 = 6/12 = 1/2
So C. This one is fast and least error prone for this problem. Thanks.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
close
