Each of four different locks has a matching key. The keys : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 16 Jan 2017, 15:22

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2795
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 226

Kudos [?]: 1619 [5] , given: 235

Each of four different locks has a matching key. The keys [#permalink]

### Show Tags

23 Sep 2010, 06:00
5
KUDOS
12
This post was
BOOKMARKED
00:00

Difficulty:

75% (hard)

Question Stats:

44% (02:15) correct 56% (01:09) wrong based on 432 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
[Reveal] Spoiler: OA

_________________

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

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html

Math Expert
Joined: 02 Sep 2009
Posts: 36520
Followers: 7066

Kudos [?]: 92914 [7] , given: 10528

Re: Each of four different locks has a matching key. The keys [#permalink]

### Show Tags

23 Sep 2010, 06:09
7
KUDOS
Expert's post
2
This post was
BOOKMARKED
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 $$P=\frac{C^2_4}{4!}=\frac{1}{4}$$.

_________________
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2795
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 226

Kudos [?]: 1619 [2] , given: 235

Re: Each of four different locks has a matching key. The keys [#permalink]

### Show Tags

23 Sep 2010, 06:14
2
KUDOS
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.

[Reveal] 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 = $$\frac{6}{24} = \frac{1}{4}$$

_________________

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

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html

Math Expert
Joined: 02 Sep 2009
Posts: 36520
Followers: 7066

Kudos [?]: 92914 [2] , given: 10528

Re: Each of four different locks has a matching key. The keys [#permalink]

### Show Tags

23 Sep 2010, 06:31
2
KUDOS
Expert's post
1
This post was
BOOKMARKED
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.

[Reveal] 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.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 36520
Followers: 7066

Kudos [?]: 92914 [1] , given: 10528

Re: Each of four different locks has a matching key. The keys [#permalink]

### Show Tags

23 Sep 2010, 06:13
1
KUDOS
Expert's post
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) $$1/2$$

Similar question with all possible scenarios: letter-arrangements-understanding-probability-and-combinats-84912.html?hilit=letter%20arrangements
_________________
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2795
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 226

Kudos [?]: 1619 [1] , given: 235

Re: Each of four different locks has a matching key. The keys [#permalink]

### Show Tags

24 Sep 2010, 06:00
1
KUDOS
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.

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

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html

Math Expert
Joined: 02 Sep 2009
Posts: 36520
Followers: 7066

Kudos [?]: 92914 [1] , given: 10528

Re: Each of four different locks has a matching key. The keys [#permalink]

### Show Tags

24 Sep 2010, 06:27
1
KUDOS
Expert's post
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.

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.
_________________
Retired Moderator
Joined: 02 Sep 2010
Posts: 805
Location: London
Followers: 105

Kudos [?]: 956 [0], given: 25

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

_________________
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2795
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 226

Kudos [?]: 1619 [0], given: 235

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

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/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: 2795
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 226

Kudos [?]: 1619 [0], given: 235

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

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html

Senior Manager
Status: Do and Die!!
Joined: 15 Sep 2010
Posts: 326
Followers: 1

Kudos [?]: 432 [0], given: 193

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: 7
Followers: 0

Kudos [?]: 0 [0], given: 46

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: 3693
Followers: 1287

Kudos [?]: 5828 [0], given: 66

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 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
_________________

Mike McGarry
Magoosh Test Prep

Senior Manager
Joined: 15 Aug 2013
Posts: 328
Followers: 0

Kudos [?]: 53 [0], given: 23

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 $$P=\frac{C^2_4}{4!}=\frac{1}{4}$$.

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: 36520
Followers: 7066

Kudos [?]: 92914 [0], given: 10528

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 $$P=\frac{C^2_4}{4!}=\frac{1}{4}$$.

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?
_________________
Current Student
Joined: 06 Sep 2013
Posts: 2035
Concentration: Finance
GMAT 1: 770 Q0 V
Followers: 62

Kudos [?]: 592 [0], given: 355

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: 36520
Followers: 7066

Kudos [?]: 92914 [0], given: 10528

Re: Each of four different locks has a matching key. The keys [#permalink]

### Show Tags

30 May 2014, 09:11
jlgdr wrote:
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

The denominator is wrong. It should be 4!, not 2^4. Please go through the solutions above.
_________________
Intern
Joined: 06 Dec 2013
Posts: 5
Followers: 0

Kudos [?]: 2 [0], given: 0

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: 65
WE: Information Technology (Computer Software)
Followers: 0

Kudos [?]: 17 [0], given: 5

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: 137
Followers: 0

Kudos [?]: 41 [0], given: 196

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

Why aren't you reducing the denominator everytime you pick?
Each of four different locks has a matching key. The keys   [#permalink] 16 Nov 2014, 13:18

Go to page    1   2    Next  [ 25 posts ]

Similar topics Replies Last post
Similar
Topics:
11 There are 5 locks and 5 keys and each of the 5 keys 3 10 Dec 2015, 10:55
5 Each of the four children has a bag with 5 different colored discs in 2 08 Aug 2015, 04:31
5 Crazy Eddie has a key chain factory. Eddie managed to 3 21 Jul 2014, 07:38
9 A key ring has 7 keys. How many different ways can they be a 8 30 Sep 2012, 16:27
18 On a game show, a contestant is given three keys, each of 7 05 Feb 2012, 16:09
Display posts from previous: Sort by