January 22, 2019 January 22, 2019 10:00 PM PST 11:00 PM PST In case you didn’t notice, we recently held the 1st ever GMAT game show and it was awesome! See who won a full GMAT course, and register to the next one. January 26, 2019 January 26, 2019 07:00 AM PST 09:00 AM PST Attend this webinar to learn how to leverage Meaning and Logic to solve the most challenging Sentence Correction Questions.
Author 
Message 
TAGS:

Hide Tags

CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2591
Location: Malaysia
Concentration: Technology, Entrepreneurship
GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35

Each of four different locks has a matching key. The keys
[#permalink]
Show Tags
23 Sep 2010, 06:00
Question Stats:
49% (01:11) correct 51% (01:01) wrong based on 707 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
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
GMAT Club Premium Membership  big benefits and savings
Gmat test review : http://gmatclub.com/forum/670to710alongjourneywithoutdestinationstillhappy141642.html




Math Expert
Joined: 02 Sep 2009
Posts: 52390

Re: Each of four different locks has a matching key. The keys
[#permalink]
Show Tags
23 Sep 2010, 06:09




Retired Moderator
Joined: 02 Sep 2010
Posts: 765
Location: London

Re: Each of four different locks has a matching key. The keys
[#permalink]
Show Tags
23 Sep 2010, 06:10
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 writeups 1) Algebra101 2) Sequences 3) Set combinatorics 4) 3D geometry
My GMAT story
GMAT Club Premium Membership  big benefits and savings



Math Expert
Joined: 02 Sep 2009
Posts: 52390

Re: Each of four different locks has a matching key. The keys
[#permalink]
Show Tags
23 Sep 2010, 06:13



CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2591
Location: Malaysia
Concentration: Technology, Entrepreneurship
GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35

Re: Each of four different locks has a matching key. The keys
[#permalink]
Show Tags
23 Sep 2010, 06:14
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 unmatching
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
GMAT Club Premium Membership  big benefits and savings
Gmat test review : http://gmatclub.com/forum/670to710alongjourneywithoutdestinationstillhappy141642.html



Math Expert
Joined: 02 Sep 2009
Posts: 52390

Re: Each of four different locks has a matching key. The keys
[#permalink]
Show Tags
23 Sep 2010, 06:31
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 unmatching
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 {Ab; Ba}). 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: {Ab; Bc; Ca} OR {Ac; Ba; Cb}. 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 the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
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. ,11 Mixed Questions, 12 Fresh Meat 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., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2591
Location: Malaysia
Concentration: Technology, Entrepreneurship
GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35

Re: Each of four different locks has a matching key. The keys
[#permalink]
Show Tags
23 Sep 2010, 06:34
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
GMAT Club Premium Membership  big benefits and savings
Gmat test review : http://gmatclub.com/forum/670to710alongjourneywithoutdestinationstillhappy141642.html



CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2591
Location: Malaysia
Concentration: Technology, Entrepreneurship
GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35

Re: Each of four different locks has a matching key. The keys
[#permalink]
Show Tags
24 Sep 2010, 06:00
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 coincidence 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
GMAT Club Premium Membership  big benefits and savings
Gmat test review : http://gmatclub.com/forum/670to710alongjourneywithoutdestinationstillhappy141642.html



Math Expert
Joined: 02 Sep 2009
Posts: 52390

Re: Each of four different locks has a matching key. The keys
[#permalink]
Show Tags
24 Sep 2010, 06:27
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 coincidence 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 the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
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. ,11 Mixed Questions, 12 Fresh Meat 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., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2591
Location: Malaysia
Concentration: Technology, Entrepreneurship
GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35

Re: Each of four different locks has a matching key. The keys
[#permalink]
Show Tags
24 Sep 2010, 07:56
Yes Bunuel, I was thinking the same but you have confirmed it. This concept is quite helpful. Thanks !!
_________________
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
GMAT Club Premium Membership  big benefits and savings
Gmat test review : http://gmatclub.com/forum/670to710alongjourneywithoutdestinationstillhappy141642.html



Senior Manager
Status: Do and Die!!
Joined: 15 Sep 2010
Posts: 268

Re: Each of four different locks has a matching key. The keys
[#permalink]
Show Tags
07 Oct 2010, 09:02
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 !!



Intern
Joined: 19 Jun 2013
Posts: 5

Re: Each of four different locks has a matching key. The keys
[#permalink]
Show Tags
15 Aug 2013, 14:17
1. to find the probability: number of needed outcomes divide by the number of all possible outcomes 2. All possible outcomes  4! 3. Find the needed number of outcomes: Two options: either 4C2= 4!/2!*2! or just write them down: FFNN(fit/non fit), FNFN, FNNF, NFFN, etc... There will be 6. therefore,
Probability= 6/4!=1/4.



Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4485

Re: Each of four different locks has a matching key. The keys
[#permalink]
Show Tags
15 Aug 2013, 14:21
Dhairya275 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?
1. 1/8 2. 1/6 3. 1/4 4. 3/8 5. 1/2 Help please ! Any Simple Solution ? Dear Dhairya275There are not hugely simple solutions to this. All solutions that occur to me involve using counting techniques. You might take a look at this blog post. http://magoosh.com/gmat/2013/gmatproba ... echniques/Total number of orders for 4 keys = 4! = 4*3*2*1 = 24 Of those 24 possible orders, how many have two keys in the right place and two in the wrong place. Suppose the locks are {a, b, c, d}. If the keys are in the order {A, B, C, D}, then all four are correct. To get two right & two wrong, we would need to select one pair from {A, B, C, D} and switch them. How many different pairs can we select from a set of four? 4C2 = \(\frac{4!}{(2!)(2!)}\) = \(\frac{4*3*2*1}{(2*1)(2*1)}\) = \(\frac{4*3}{2}\) = 6 So, of the 24 sets, 6 of them would have two right & two wrong. P = 6/24 = 1/4 Does this make sense? Mike
_________________
Mike McGarry Magoosh Test Prep
Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)



Manager
Joined: 15 Aug 2013
Posts: 247

Re: Each of four different locks has a matching key. The keys
[#permalink]
Show Tags
21 Apr 2014, 18:13
Bunuel wrote: 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 # of ways to assign the keys to the locks is \(4!\). \(C^2_4\) to choose which 2 keys will fit. Other 2 keys can be aaranged only one way. So \(P=\frac{C^2_4}{4!}=\frac{1}{4}\). Answer: C. I'm definitely missing something rather fundamental because I can't make out what connection I'm missing. I'm not able to apply the correct methodology to the correct problem  ever. The way I tackled this problem(which was obviously wrong was): There are 4 keys and 4 locks, therefore 16 combinations? Out of 16 possibilities, there are only 2 ways to arrange this so that the locks match  (Lock A, Key A), (LB, KB), (LC,KD) and (LD, KC) and another way is LBKB, LAKA, LCKD, LDKC. Therefore, I calculated 2/16 which is 1/8.



Math Expert
Joined: 02 Sep 2009
Posts: 52390

Re: Each of four different locks has a matching key. The keys
[#permalink]
Show Tags
22 Apr 2014, 00:58
russ9 wrote: Bunuel wrote: 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 # of ways to assign the keys to the locks is \(4!\). \(C^2_4\) to choose which 2 keys will fit. Other 2 keys can be aaranged only one way. So \(P=\frac{C^2_4}{4!}=\frac{1}{4}\). Answer: C. I'm definitely missing something rather fundamental because I can't make out what connection I'm missing. I'm not able to apply the correct methodology to the correct problem  ever. The way I tackled this problem(which was obviously wrong was): There are 4 keys and 4 locks, therefore 16 combinations? Out of 16 possibilities, there are only 2 ways to arrange this so that the locks match  (Lock A, Key A), (LB, KB), (LC,KD) and (LD, KC) and another way is LBKB, LAKA, LCKD, LDKC. Therefore, I calculated 2/16 which is 1/8. A  B  C  D (locks) a  b  c  d (keys) a  b  d  c a  c  b  d a  c  d  b a  d  b  c a  d  c  b b  a  c  d b  a  d  c b  c  a  d b  c  d  a b  d  a  c b  d  c  a c  a  b  d c  a  d  b c  b  a  d c  b  d  a c  d  a  b c  d  b  a d  a  b  c d  a  c  b d  b  a  c d  b  c  a d  c  a  b d  c  b  a As you can see there are 4!=24 ways to assign keys to locks (arrangement of 4 keys). So, the number of total outcomes is 24. Next, consider a case when exactly two of the keys fit the locks, say a and b fit and d and c not: A  B  C  D (locks) a  b  d  c (keys) As you can see if a and b fit, d and c not to fit can only be arranged in one way as shown above. Now, the number of ways to choose which 2 keys fit is \(C^2_4\), so the number of favorable outcomes is \(1*C^2_4=6\). P = 6/24 = 1/4. Does this make sense?
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
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. ,11 Mixed Questions, 12 Fresh Meat 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., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



SVP
Joined: 06 Sep 2013
Posts: 1705
Concentration: Finance

Re: Each of four different locks has a matching key. The keys
[#permalink]
Show Tags
30 May 2014, 08:55
I think I got this one wrong.
I did the following
4C2 * (1/2)^4
4C2: Choosing which 2 keys will be the correct keys
(1/2)^4 = 1 correct and one incorrect choice for each key
Therefore, 3/8
Could anyone suggest what's wrong with this method?
Thanks Cheers J



Math Expert
Joined: 02 Sep 2009
Posts: 52390

Re: Each of four different locks has a matching key. The keys
[#permalink]
Show Tags
30 May 2014, 09:11



Intern
Joined: 06 Dec 2013
Posts: 5

Re: Each of four different locks has a matching key. The keys
[#permalink]
Show Tags
21 Jun 2014, 03:52
Just another approach. The probability of choosing the first matching key is 1/4, the probability of choosing the second matching key is 1/3, the probability of choosing the third not matching key is 1/2 and the probability of choosing the fourth not matching key is 1. We've got 4!/2!2! =6 ways of doing so. Thus the answer is 6*1/4*1/3*1/2*1 = 6/24 = 1/4



Manager
Joined: 24 Oct 2012
Posts: 63
WE: Information Technology (Computer Software)

Re: Each of four different locks has a matching key. The keys
[#permalink]
Show Tags
21 Jun 2014, 23:34
I was just going over probability questions. can some one explain me what's wrong in my approach here.
Probability of choosing one right key out of 4 is 1/4.
Probablity of choosing another right key is 1/4.
since the question is asking for 2 right keys , probability is multiplication of both = 1/4 * 1/4 = 1/16.
I went through explanations here. but this is how I solved when i looked at problem. Can some one correct me why is this approach not taken?
Thanks



Manager
Joined: 30 Mar 2013
Posts: 109

Each of four different locks has a matching key. The keys
[#permalink]
Show Tags
16 Nov 2014, 13:18
Chance of getting the first key to match is 1/4 Second key 1/3 Third key DOESN'T match is 1/2 last key doesn't match =1/1 or 1
In how many ways can you accomplish this?4C2 = 6
Therefore the answer is 1/4
Why aren't you reducing the denominator everytime you pick?




Each of four different locks has a matching key. The keys &nbs
[#permalink]
16 Nov 2014, 13:18



Go to page
1 2
Next
[ 32 posts ]



