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Manager  Joined: 10 Mar 2013
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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]

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

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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

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

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FIT FIT NOFIT NOFIT
$$(1/4) * (1/3) * (1/2) * 1 * \frac{4!}{(2!2!)} = 1/4$$
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GMAT 1: 690 Q49 V35 Re: Each of four different locks has a matching key. The keys  [#permalink]

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

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

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

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

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

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

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

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

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

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

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Bunuel Hi, I don't understand why the number of desired outcomes is 6 in this question. So we need to find arrangements in which 2 locks have their right keys. According to me, the desired outcomes should 12 because for example, the pair of locks with right keys is L1 and L2. Therefore, one arrangement can be L1, L2, L3, L4 but we can also have an arrangement L2, L1, L3, L4.
Similarly, the number of desired arrangements taking a pair into consideration should be multiplied by 2. Therefore, the number of desired arrangements should be 2C4 * 2.
Am I missing something?
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Re: Each of four different locks has a matching key. The keys  [#permalink]

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Hi...
I used the concept of events happening r times out of n times...nCr P^r (1-P)^(n-r)...=4C2 (.5)^2 (.5)^2

Please explain where did I get it wrong
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Re: Each of four different locks has a matching key. The keys  [#permalink]

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1
Debashis Roy wrote:

Hi...
I used the concept of events happening r times out of n times...nCr P^r (1-P)^(n-r)...=4C2 (.5)^2 (.5)^2

Please explain where did I get it wrong

You cannot use this formula, because the probability has to follow some restriction, that is two are correctly used. If two are correctly used, then the other two are wrong.
So the last two are connected to the first two..

If you had a question about coin, and we look for two heads, then we use the method you have discussed, because each is independent of other throws.
But that is not the case here.
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Re: Each of four different locks has a matching key. The keys  [#permalink]

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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 P of keys ; 1/4*1/3*1/2 *1/1 = 1/24
and for 2 locks 4c2 = 6
1/24 * 6 = 1/4
IMO C
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Re: Each of four different locks has a matching key. The keys  [#permalink]

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Bunuel wrote:
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.

help me out Bunnuel,

4 Locks 4 Keys

Total combinations: 16

first gets right: 1/16
2nd gets right: 1/15
3rd gets wrong: 13/14
4th gets wrong: 12/13

1/16x1/15x13/14x12/13 x 4! = why this is not correct.

can you help me out with this method?
I want to know where I missed...
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Re: Each of four different locks has a matching key. The keys  [#permalink]

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1
u can make cases. any of the two locks can take the right key hence u have 6 cases(1,2)(1,3) (1,4) (2,3) (2,4) (3,4): here 1,2,3,4 represents locks. 4 keys can be arranged in 4! ways. thus probability=6/4!=1/4 Re: Each of four different locks has a matching key. The keys   [#permalink] 05 Apr 2019, 00:18

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