GMAT Changed on April 16th - Read about the latest changes here

It is currently 22 May 2018, 10:52

Close

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

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Each of four different locks has a matching key. The keys

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Manager
Manager
User avatar
Joined: 10 Mar 2013
Posts: 243
GMAT 1: 620 Q44 V31
GMAT 2: 690 Q47 V37
GMAT 3: 610 Q47 V28
GMAT 4: 700 Q50 V34
GMAT 5: 700 Q49 V36
GMAT 6: 690 Q48 V35
GMAT 7: 750 Q49 V42
GMAT 8: 730 Q50 V39
Each of four different locks has a matching key. The keys [#permalink]

Show Tags

New post 17 Jul 2015, 19:30
Bunuel, is this valid?
p(2 keys correct)=Match*Match*Miss*Miss=(1/4)*(1/3)*(1/2)*1 = 1/24
P(2 keys correct)=p(2 keys correct)*Total combinations that this event can happen
Total combinations that this event can happen = Number of attempted matches that will be successful = Number of ways we can arrange Ma,Ma,Mi,Mi = C(4,2)
P(2 keys correct)=1/24*C(4,2)=1/24*6=1/4
Expert Post
SVP
SVP
User avatar
P
Joined: 08 Jul 2010
Posts: 2099
Location: India
GMAT: INSIGHT
WE: Education (Education)
Reviews Badge
Each of four different locks has a matching key. The keys [#permalink]

Show Tags

New post 18 Jul 2015, 06:49
Quote:
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


TooLong150 wrote:
Bunuel, is this valid?
p(2 keys correct)=Match*Match*Miss*Miss=(1/4)*(1/3)*(1/2)*1 = 1/24
P(2 keys correct)=p(2 keys correct)*Total combinations that this event can happen
Total combinations that this event can happen = Number of attempted matches that will be successful = Number of ways we can arrange Ma,Ma,Mi,Mi = C(4,2)
P(2 keys correct)=1/24*C(4,2)=1/24*6=1/4


Hi TooLong150,

Your explanation is ABSOLUTELY CORRECT :)

Just another Traditional way of solving the same question

\(Probability = \frac{Favourable Outcomes}{Total Outcomes}\)

Total Possible Arrangement of 4 keys to 4 locks = 4! = 24

Favourable Ways of Arrangement of Keys (2 correct and 2 incorrect) = 4C2*1 = 6

4C2 represents the ways to select the lock in which the key is found to be correct
1 is the way to arrange the remaining two keys to remaining two locks such that key doesn't match the lock


i.e. Probability = 6/24 = 1/4

Answer: Option C
_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Manager
Manager
avatar
Joined: 07 Apr 2015
Posts: 177
Each of four different locks has a matching key. The keys [#permalink]

Show Tags

New post 02 Aug 2015, 01:48
FIT FIT NOFIT NOFIT
\((1/4) * (1/3) * (1/2) * 1 * \frac{4!}{(2!2!)} = 1/4\)
Current Student
avatar
Joined: 15 May 2015
Posts: 39
GMAT 1: 690 Q49 V35
GMAT ToolKit User
Re: Each of four different locks has a matching key. The keys [#permalink]

Show Tags

New post 04 Aug 2015, 06:03
Hi Bunuel ,

If both keys and locks are different,then isnt the total number of ways is 4^2(2 keys and 4 locks).
and the number of desired cases is 2^2.

so probability = 1/4

Can you please check whether the approach is right.
Board of Directors
User avatar
P
Joined: 17 Jul 2014
Posts: 2735
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)
GMAT ToolKit User Premium Member Reviews Badge
Re: Each of four different locks has a matching key. The keys [#permalink]

Show Tags

New post 26 Mar 2016, 17:51
my approach...
we need to select 2 keys out of 4 to be chosen correctly. we can get this in 4C2 ways, or 6 ways.
now..we have 4 doors...
to unlock the first one - 1/4
to unlock the second one - 1/3
we are left with 2 keys..we need to make sure that they do not match
probability that one key doesn't match 1/2
probability that the other one doesn't match 1.
now..probability that 2 are unlocked and 2 not:
1/4 * 1/3 * 1/2 * 1 = 1/24
since there are 6 possible arrangements, multiply 1/24 with 6.
1/4

C
Director
Director
User avatar
P
Joined: 26 Aug 2016
Posts: 572
Location: India
Concentration: Strategy, Marketing
GMAT 1: 690 Q50 V33
GMAT 2: 700 Q50 V33
GPA: 4
WE: Consulting (Consulting)
Premium Member CAT Tests
Re: Each of four different locks has a matching key. The keys [#permalink]

Show Tags

New post 01 May 2017, 10:28
mikemcgarry wrote:
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 Dhairya275
There 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/gmat-proba ... 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 :-)


Thank you . It helped me a lot . :) .
Intern
Intern
avatar
B
Joined: 20 Aug 2017
Posts: 16
GMAT ToolKit User
Re: Each of four different locks has a matching key. The keys [#permalink]

Show Tags

New post 13 Dec 2017, 12:14
Bunuel wrote:
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.

Spoiler: :: My Solution
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.


Hey Bunuel,

can you please tell me if theres any article that explains this C technique in the forum?

thank you!
Intern
Intern
avatar
B
Joined: 01 Dec 2017
Posts: 14
Location: Italy
Schools: IMD Jan'18
GMAT 1: 680 Q46 V38
GPA: 4
Re: Each of four different locks has a matching key. The keys [#permalink]

Show Tags

New post 13 Dec 2017, 13:02
GMatAspirerCA wrote:
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


You have to remember that the four events ar NOT independent, therefore you can't do 1/4/*1/4....Once you got the first key right probability of the second to be right becomes 1/3
Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 45248
Re: Each of four different locks has a matching key. The keys [#permalink]

Show Tags

New post 13 Dec 2017, 20:15
omavsp wrote:
Bunuel wrote:
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.

Spoiler: :: My Solution
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.


Hey Bunuel,

can you please tell me if theres any article that explains this C technique in the forum?

thank you!



21. Combinatorics/Counting Methods



For more:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread


Hope it helps.
_________________

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?
Extra-hard Quant Tests with Brilliant Analytics

Intern
Intern
avatar
B
Joined: 01 Dec 2017
Posts: 14
Location: Italy
Schools: IMD Jan'18
GMAT 1: 680 Q46 V38
GPA: 4
Re: Each of four different locks has a matching key. The keys [#permalink]

Show Tags

New post 14 Dec 2017, 06:21
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.


Hi Bunuel ,

How would the solution change if the question was "probability of not getting ANY keys in the right lock?"

I have found the following solution by enumeration, but I can't find a more analytical approach to solve this....

First we can choose three keys to couple with lock A in the wrong way : b,c,d
once we chose the first key, we can always find three ways to NOT fit any of the remaining keys; for example If we choose key "b" for the first lock we will have
A B C D
b a d c
b d a c
b c d a

Therefore we have \(3*3=9\) ways to get all the keys "wrong". I think this formula can be interpreted as 3* \(C^3_2\) , but I am not convinced as why....may be it's like "in how many ways we can choose 2 keys wrong out of three"? It is also a bit strange cause in this case we already know that one key will never fit in any case....they are clearly a part of all the possible permutation, but I would like to find a more academic way to solve it....


Could you help me out with this please , Bunuel?
Many many thanks!!
Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 45248
Re: Each of four different locks has a matching key. The keys [#permalink]

Show Tags

New post 14 Dec 2017, 06:24
teone83 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.


Hi Bunuel ,

How would the solution change if the question was "probability of not getting ANY keys in the right lock?"

I have found the following solution by enumeration, but I can't find a more analytical approach to solve this....

First we can choose three keys to couple with lock A in the wrong way : b,c,d
once we chose the first key, we can always find three ways to NOT fit any of the remaining keys; for example If we choose key "b" for the first lock we will have
A B C D
b a d c
b d a c
b c d a

Therefore we have \(3*3=9\) ways to get all the keys "wrong". I think this formula can be interpreted as 3* \(C^3_2\) , but I am not convinced as why....may be it's like "in how many ways we can choose 2 keys wrong out of three"? It is also a bit strange cause in this case we already know that one key will never fit in any case....they are clearly a part of all the possible permutation, but I would like to find a more academic way to solve it....


Could you help me out with this please , Bunuel?
Many many thanks!!


Check here: https://gmatclub.com/forum/letter-arran ... 84912.html
_________________

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?
Extra-hard Quant Tests with Brilliant Analytics

Re: Each of four different locks has a matching key. The keys   [#permalink] 14 Dec 2017, 06:24

Go to page   Previous    1   2   [ 31 posts ] 

Display posts from previous: Sort by

Each of four different locks has a matching key. The keys

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


cron

GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.