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

Question

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

Bunuel · Accepted Answer

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.

Bunuel · Answer

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.

shrouded1 · Answer

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

So probability = 6/24 = 1/4

Answer (c)

Bunuel · Answer

Similar question with all possible scenarios: letter-arrangements-understanding-probability-and-combinats-84912.html?hilit=letter%20arrangements

gurpreetsingh · Answer

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}

gurpreetsingh · Answer

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

gurpreetsingh · Answer

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.

Bunuel · Answer

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.

gurpreetsingh · Answer

Yes Bunuel, I was thinking the same but you have confirmed it. This concept is quite helpful.

Thanks !!

mikemcgarry · Answer

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. https://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 :-)

russ9 · Answer

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.

Bunuel · Answer

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?

TooLong150 · Answer

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

GMATinsight · Answer

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

GmatPoint · Answer

The ways of selecting two keys right = 4C2
The total combinations possible = 4!
Thus, probability = 4C2/4! = 4*3/2*4!
 = 1/4.

Thus, the correct option is C.

TanushreeLahoti · Answer

What if we take Miss*Miss*Match*Match - (3/4)*(2/3)*(1/2)*1 = 1/4 ?
As per this, final answer is not coming to 1/4. What am I missing?

PrinceDAYAL · Answer

Hello  Bunuel ,

I tried to do this as followed, Can you please tell where did I go wrong?
4 keys, 4 locks
P(Success = When key matches) = 1/4
P(Failure = Key doesn't matches) = 3/4

Now for exactly 2 to be successful other should not be 
4C2 * (1/4)^2 * (3/4)^2

Don't know what went wrong in this solution?

NilayMaheshwari · Answer

Let us denote matching by M and not matching by N

1 key matching - MNNN- No. of ways to arrange = 4!/3! = 4 ways
2 keys matching- MMNN- No. of ways to arrange = 4!/2!2! = 6 ways
3 keys matching- MMMN- No. of ways to arrange same as 1 = 4 ways
4 keys matching- MMMM- No.of ways to arrange = 1
None of the keys matching- No.of ways to arrange = 9
Ways to arrange where none of the keys are matching can be as below-
BCDA CADB DABC
BDAC CDBA DCAB
BADC CDAB DCBA

Total number of ways = 4+6+4+1+9 = 24 ways
No. of ways to arrange where two keys are matching = 6 ways

Probability = 6/24 = 1/4.

I couldn't find an easier approach to count the number of ways to arrange none of the keys matching part. Maybe someone can help.

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

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