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

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23 Sep 2010, 07:00

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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?

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

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23 Sep 2010, 07:09

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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 # 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.

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

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23 Sep 2010, 07:14

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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.

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

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23 Sep 2010, 07:31

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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.

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 \(C^2_5*2\).

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

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23 Sep 2010, 07:13

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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?

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

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24 Sep 2010, 07:00

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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. _________________

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

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24 Sep 2010, 07:27

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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.

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

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23 Sep 2010, 07:10

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]

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

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23 Sep 2010, 07:34

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 _________________

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

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07 Oct 2010, 10:02

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. _________________

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

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15 Aug 2013, 15:17

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,

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

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15 Aug 2013, 15:21

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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?

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

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21 Apr 2014, 19:13

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.

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

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22 Apr 2014, 01:58

Expert's post

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 \(1*C^2_4=6\).

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

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21 Jun 2014, 04:52

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

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