|
Author |
Message |
|
TAGS:
|
|
|
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2762
Location: Malaysia
Concentration: Marketing, Entrepreneurship
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 126
Kudos [?]:
644
[1] , given: 222
|
Each of four different locks has a matching key. The keys [#permalink]
 23 Sep 2010, 07:00
1
This post received
KUDOS
Question Stats:
47% (02:18) correct
52% (01:21) wrong based on 21 sessions
Each of four different locks has a matching key. The keys are randomly reassigned to the locks. What is the probability that exactly two of the keys fit the locks to which they are reassigned? A 1/8B. 1/6C. 1/4D. 3/8E. 1/2
_________________
Fight for your dreams :For all those who fear from Verbal- lets give it a fight
Money Saved is the Money Earned 
Jo Bole So Nihaal , Sat Shri Akaal
 Support GMAT Club by putting a GMAT Club badge on your blog/Facebook
Find out what's new at GMAT Club - latest features and updates
Gmat test review :
670-to-710-a-long-journey-without-destination-still-happy-141642.html
|
|
|
|
|
|
|
GMAT Club team member
Joined: 02 Sep 2009
Posts: 12116
Followers: 1879
Kudos [?]:
10132
[1] , given: 965
|
1
This post received
KUDOS
|
|
|
|
|
|
GMAT Club team member
Joined: 02 Sep 2009
Posts: 12116
Followers: 1879
Kudos [?]:
10132
[1] , given: 965
|
1
This post received
KUDOS
|
|
|
|
|
|
GMAT Club team member
Joined: 02 Sep 2009
Posts: 12116
Followers: 1879
Kudos [?]:
10132
[1] , given: 965
|
1
This post received
KUDOS
gurpreetsingh wrote: This is first question of Mgmat challenge set of Gmat Club tests. Though while solving the tests my question was wrong, but later I tried to solve it and got the correct answer. I looked at the explanation it was too long. I have done in a simple way, but I m not 100% if I m correct. OA is C -1/4
Let L1,L2,L3,L4 are locks with K1,K2,K3,K4 respective keys.
Final output after merging them is L1K1, L2K2, L3K4, L4K3.
Now we have to find the probability of happening the above arrangement.
What I did was, I supposed the above arrangement to be M M U U where M - matching, U- un-matching
The above can be arranged in 4!/2!2! = 6
Total number of arrangement = total number of ways = 4! = 24
Hence the probability = \frac{6}{24} = \frac{1}{4} No this approach is not right (you've got the correct answer because 2 keys which should be assigned incorrectly can be assigned only in 1 way {A-b; B-a}). Consider this: if it were 5 locks instead of 4 and everything else remained the same. Your approach would give MMUUU = \frac{5!}{2!3!}=10 --> total # of assignments 5! --> P=\frac{10}{120}. But correct answer would be: C^2_5 - choosing which 2 keys will fit --> other 3 keys can be arranged so that no other key to fit in 2 ways: {A-b; B-c; C-a} OR {A-c; B-a; C-b}. So total # of ways to assign exactly 2 keys to fit would be C^2_5*2. So P=\frac{C^2_5*2}{5!}=\frac{20}{120}. Hope it's clear.
_________________
NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!
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; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders
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!!! ,11 Mixed Questions NEW!!!, 12 Fresh Meat 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!!!, 11 New DS set. NEW!!!
What are GMAT Club Tests?
25 extra-hard Quant Tests
Find out what's new at GMAT Club - latest features and updates
|
|
|
|
|
|
Retired Moderator
Joined: 02 Sep 2010
Posts: 814
Location: London
Followers: 57
Kudos [?]:
307
[0], given: 25
|
gurpreetsingh wrote: Each of four different locks has a matching key. The keys are randomly reassigned to the locks. What is the probability that exactly two of the keys fit the locks to which they are reassigned?
a) 1/8
b) 1/6
c) 1/4
d) 3/8
e) 1/2 Total ways to assign keys = 4! = 24 Ways to assign keys such that only 2 fit = Choose the 2 that fit = C(4,2) = 6 Note that once you pick the two locks on which the keys fit there is exactly one allocation of keys possible. For eg. If you pick Lock A & Lock B fit, the only allocation possible is [Key A, Key B, Key D, Key C] So probability = 6/24 = 1/4 Answer (c)
_________________
Math write-ups
1) Algebra-101 2) Sequences 3) Set combinatorics 4) 3-D geometry
My GMAT story
Find out what's new at GMAT Club - latest features and updates
|
|
|
|
|
|
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2762
Location: Malaysia
Concentration: Marketing, Entrepreneurship
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 126
Kudos [?]:
644
[0], given: 222
|
This is first question of Mgmat challenge set of Gmat Club tests. Though while solving the tests I got it wrong, but later I tried to solve it and got the correct answer. I looked at the explanation it was too long. I have done in a simple way, but I m not 100% if I m correct. OA is C -1/4
Let L1,L2,L3,L4 are locks with K1,K2,K3,K4 respective keys.
Final output after merging them is L1K1, L2K2, L3K4, L4K3.
Now we have to find the probability of happening the above arrangement.
What I did was, I supposed the above arrangement to be M M U U where M - matching, U- un-matching
The above can be arranged in 4!/2!2! = 6
Total number of arrangement = total number of ways = 4! = 24
Hence the probability = \frac{6}{24} = \frac{1}{4} _________________
Fight for your dreams :For all those who fear from Verbal- lets give it a fight
Money Saved is the Money Earned
Jo Bole So Nihaal , Sat Shri Akaal
Support GMAT Club by putting a GMAT Club badge on your blog/Facebook
Find out what's new at GMAT Club - latest features and updates
Gmat test review :
670-to-710-a-long-journey-without-destination-still-happy-141642.html
|
|
|
|
|
|
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2762
Location: Malaysia
Concentration: Marketing, Entrepreneurship
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 126
Kudos [?]:
644
[0], given: 222
|
Thanks Bunnel I got it.... Your letter arrangement thread is quite good !! I m able to answer probability questions, but these letter arrangement sometimes confuses me. +1
_________________
Fight for your dreams :For all those who fear from Verbal- lets give it a fight
Money Saved is the Money Earned
Jo Bole So Nihaal , Sat Shri Akaal
Support GMAT Club by putting a GMAT Club badge on your blog/Facebook
Find out what's new at GMAT Club - latest features and updates
Gmat test review :
670-to-710-a-long-journey-without-destination-still-happy-141642.html
|
|
|
|
|
|
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2762
Location: Malaysia
Concentration: Marketing, Entrepreneurship
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 126
Kudos [?]:
644
[0], given: 222
|
A basket contains 3 white and 5 blue balls. Mary will extract one ball at random and keep it. If, after that, John will extract one ball at random, what is the probability that John will extract a blue ball? The above question is from M14 q1. I solved it using steps by taking different conditions and solved it correctly. the answer was 5/8. I was wondering is this true for probability that whatever happens in between, the finally probability will remain same? if I exclude "Mary will extract one ball at random and keep it. If, after that," still the probability is 5/8. is this a co-incidence for a particular case? I know if it had mentioned that mary picked a particular ball, then probability would have changed. I m talking about the case in which no specific thing mentioned. Actually I have seen such question earlier and I though it would be better to confirm it. I think if mary picks 2 balls without mentioning which ones, then the probability will remain 5/8.
_________________
Fight for your dreams :For all those who fear from Verbal- lets give it a fight
Money Saved is the Money Earned
Jo Bole So Nihaal , Sat Shri Akaal
Support GMAT Club by putting a GMAT Club badge on your blog/Facebook
Find out what's new at GMAT Club - latest features and updates
Gmat test review :
670-to-710-a-long-journey-without-destination-still-happy-141642.html
|
|
|
|
|
|
GMAT Club team member
Joined: 02 Sep 2009
Posts: 12116
Followers: 1879
Kudos [?]:
10132
[0], given: 965
|
gurpreetsingh wrote: A basket contains 3 white and 5 blue balls. Mary will extract one ball at random and keep it. If, after that, John will extract one ball at random, what is the probability that John will extract a blue ball?
The above question is from M14 q1. I solved it using steps by taking different conditions and solved it correctly.
the answer was 5/8.
I was wondering is this true for probability that whatever happens in between, the finally probability will remain same?
if I exclude "Mary will extract one ball at random and keep it. If, after that," still the probability is 5/8.
is this a co-incidence for a particular case?
I know if it had mentioned that mary picked a particular ball, then probability would have changed. I m talking about the case in which no specific thing mentioned. Actually I have seen such question earlier and I though it would be better to confirm it. I think if mary picks 2 balls without mentioning which ones, then the probability will remain 5/8. The initial probability of drawing blue ball is 5/8. Without knowing the other results , the probability of drawing blue ball will not change for ANY successive drawing: second, third, fourth... There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results). If Mary extracts not 1 but 7 balls at random and keep them, then even after that, the probability that the last ball left for John is blue still will be 5/8. Hope it's clear.
_________________
NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!
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; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders
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!!! ,11 Mixed Questions NEW!!!, 12 Fresh Meat 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!!!, 11 New DS set. NEW!!!
What are GMAT Club Tests?
25 extra-hard Quant Tests
Find out what's new at GMAT Club - latest features and updates
|
|
|
|
|
|
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2762
Location: Malaysia
Concentration: Marketing, Entrepreneurship
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 126
Kudos [?]:
644
[0], given: 222
|
|
|
|
|
|
|
Senior Manager
Status: Do and Die!!
Joined: 15 Sep 2010
Posts: 335
Followers: 1
Kudos [?]:
27
[0], given: 192
|
MGMT 1 probability [#permalink]
07 Oct 2010, 10:02
Each of four different locks has a matching key. The keys are randomly reassigned to the locks. What is the probability that exactly two of the keys fit the locks to which they are reassigned? 1/8 1/6 1/4 3/8 1/2 Each of four locks has a matching key => 4 original keys and 4 matching keys. => total 8 keys. probability of finding 2 keys. => 2/8=>1/4=> correct..... i don't know if my approach is right or wrong. Please comment thanks.
_________________
I'm the Dumbest of All !!
|
|
|
|
|
|
|
MGMT 1 probability
[#permalink]
07 Oct 2010, 10:02
|
|
|
|
|
|
|
|
|
|
|
close
