ps from gmat club

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ps from gmat club

by Ramit88 » Thu Jan 13, 2011 8:55 am
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 1/8



ans c


my approch
prob(right key)= 1/4
prob(wrong key)= 1-1/4
= 3/4

the probability that exactly two of the keys fit the locks to which they are reassigned= 1/2 x 1/2 x 3/4x 3/4
9/64
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by fskilnik@GMATH » Thu Jan 13, 2011 9:33 am
Hi there, Ramit!

Let L1, L2, L3 and L4 be the 4 different locks and K1, K2, K3 and K4 the corresponding (matching) keys!

The total number of possibilities to randomly choose the keys to the locks is P4 = 4!, because you have 4 keys for L1, then 3 keys for L2, then 2 keys for L3 and then one (the last) key for L4.

How many of the 4! (equiprobable) possibilities mentioned above are favorable?

Exactly the number of choices of 2 of the 4 locks (to receive their proper matching keys): C(4,2) = 6 possibilities.

Explanation: one of the 6 possibilities is L1 and L2, meaning the following full scenario: (L1,K1) ; (L2,K2) ; (L3, K4) and (L4, K3) , because we want EXACTLY two right matches, remember that!

Therefore the answer to the problem is 6/4! = 1/4 .

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by Anurag@Gurome » Thu Jan 13, 2011 9:43 am
Ramit88 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 1/8
If two of the four keys are assigned to their matching locks, there is two possible arrangements for the other two keys. In one of them they are also assigned to their matching locks and in the other one they are not. Thus the number of possible arrangements such that exactly two of the keys are assigned correctly is the same as the number of possible selections of two keys out of four.

Hence total number of such arrangements = 4C2 = 6
And total number of possible arrangements = 4! = 24

Required probability = 6/24 = 1/4

The correct answer is C.
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by Ramit88 » Thu Jan 13, 2011 9:50 am
thank u Anurag and Fabio..

but can u pls tell why my approach is wrong..

my approach
total keys =4
prob(right key)= 1/4
prob(wrong key) = 1-1/4
= 3/4

the probability that exactly two of the keys fit the locks to which they are reassigned= p(Right key) x p(Right key) x p(Wrong key) x p(Wrong key)
=1/2 x 1/2 x 3/4x 3/4
=9/64

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by Anurag@Gurome » Thu Jan 13, 2011 9:57 am
Ramit88 wrote:my approach
total keys =4
prob(right key)= 1/4
prob(wrong key) = 1-1/4
= 3/4

the probability that exactly two of the keys fit the locks to which they are reassigned= p(Right key) x p(Right key) x p(Wrong key) x p(Wrong key)
=1/2 x 1/2 x 3/4x 3/4
=9/64
You can multiply probability of some events to get the joint probability of those events only when the events are independent. But in this case, the correct arrangement exactly two key also results in wrong arrangement other two. Thus wrong arrangement of remaining two key depends on the right arrangement of first two keys.

While dealing with probability of dependent events, you have to select your methods accordingly.

Hope it helps.
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by fskilnik@GMATH » Thu Jan 13, 2011 9:59 am
Anurag@Gurome wrote:If two of the four keys are assigned to their matching locks, there is two possible arrangements for the other two keys. In one of them they are also assigned to their matching locks and in the other one they are not.
When you start this sentence you admit two keys are assigned to their matching locks, therefore you are starting the counting of FAVORABLE scenarios. That means you must focus in EXACTLY two matches, what is to say that the arrangements "for the other two keys" (your words) should "not match", and that means that there is just one arrangement "for the other two keys" for every (one of the six) choices of the two keys assigned to their matching locks.

I know you have that in mind, but your explanation does not make that clear. Please pay attention to the part of my explanation in bold.

Regards,
Fabio.
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by fskilnik@GMATH » Thu Jan 13, 2011 10:15 am
Ramit88 wrote:thank u Anurag and Fabio..

but can u pls tell why my approach is wrong..

my approach
total keys =4
prob(right key)= 1/4
prob(wrong key) = 1-1/4
= 3/4

the probability that exactly two of the keys fit the locks to which they are reassigned= p(Right key) x p(Right key) x p(Wrong key) x p(Wrong key)
=1/2 x 1/2 x 3/4x 3/4
=9/64
Hi there, Ramit.

Prob(right key) = 1/4 and Prob(wrong key) = 3/4 are correct in the sense that CHOSEN one of the locks, the probability to choose the right key [wrong key] to it is really 1/4 [3/4].

Now You CAN say the following:

P(L1 gets K1) = 1/4 , then P(L2 gets K2) = 1/3 because you have keys K2, K3 and K4 to choose to L2, and just K2 fits.

Now you have to be careful, because you want P(L3 gets K4) = 1/2 and P(L4 gets K3) = 1/1 (it´s the last choice of key to the last lock... no choices to be honest), therefore the full scenario (L1,K1), (L2, K2), (L3, K4) and (L4,K3) has probability 1/4 * 1/3 * 1/2 * 1 = 1/24

Now what? You could calculate exactly the same probability for exactly other 2 locks to get the proper keys, etc... there are C(4,2) = 6 different scenarios, all of them equiprobable and mutually exclusive, therefore we get: 1/24 + .... + 1/24 = 1/4

Please compare my approach to yours, because I guess you have tried to do that without success.

Regards,
Fabio.
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by Anurag@Gurome » Thu Jan 13, 2011 10:31 am
fskilnik wrote:When you start this sentence ...
I know you have that in mind, but your explanation does not make that clear. Please pay attention to the part of my explanation in bold.
Thanks Fabio.
But I didn't started the sentence to calculate the number of favorable outcomes. I mentioned that "If two of the four keys are assigned to their matching locks, there is two possible arrangements for the other two keys. In one of them they are also assigned to their matching locks and in the other one they are not". I didn't mentioned anything about "exactly two keys go to their correct places". I've just stated the possibilities here given the fact that already two keys have already assigned to their correct places.

In the second sentence I introduced the fact that "exactly" two keys going to their correct places is possible in only one arrangement given the fact that already two keys have already assigned to their correct places.

Hope I'm more clear now.
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by fskilnik@GMATH » Thu Jan 13, 2011 10:45 am
Hi there, Anurag.

I´m glad you didn´t feel upset with my comment. Please feel free to do the same whenever you believe necessary.
Anurag@Gurome wrote:I didn't mentioned anything about "exactly two keys go to their correct places". I've just stated the possibilities here given the fact that already two keys have already assigned to their correct places.
I know, that´s why I didn´t say it was wrong. I just meant that when the sentence started with two keys going to proper places, the reader will believe you are dealing with the favorable scenario, because we are always looking for "possible x desirable" cases, therefore if you are not dealing with one of them, people will believe you are dealing with the other.

As far as clearness is concerned, I believe:
Anurag@Gurome wrote: Thus the number of possible arrangements such that exactly two of the keys are assigned correctly is the same as the number of possible selections of two keys out of four.
is not as clear as:
Anurag@Gurome wrote: ... "exactly" two keys going to their correct places is possible in only one arrangement given the fact that already two keys have already assigned to their correct places.
and this is not as clear as
... "exactly" two keys going to their correct places is possible in only one arrangement given the fact that already two keys have already assigned to their correct places, therefore there is just one possible fit to the other two locks: get the last two keys "wrong".
Regards,
Fabio.
Last edited by fskilnik@GMATH on Thu Jan 13, 2011 10:48 am, edited 1 time in total.
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by Anurag@Gurome » Thu Jan 13, 2011 10:47 am
Thanks Fabio.
Got your point. :)
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by fskilnik@GMATH » Thu Jan 13, 2011 10:50 am
Anurag@Gurome wrote:Thanks Fabio.
Got your point. :)
Cheers! See you in other BTG posts! :)
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by GMATGuruNY » Thu Jan 13, 2011 12:12 pm
Ramit88 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 1/8



ans c


my approch
prob(right key)= 1/4
prob(wrong key)= 1-1/4
= 3/4

the probability that exactly two of the keys fit the locks to which they are reassigned= 1/2 x 1/2 x 3/4x 3/4
9/64
Another approach:

To determine the probability that an event happens exactly n times:

P(exactly n times) = P(one way) * (total possible ways)

For the problem above, we want to determine the probability that exactly 2 keys are correctly assigned. One way for exactly 2 keys to be correctly assigned is for the first 2 keys to be correct and for the last 2 keys to be wrong:

P(1st key is correct) = 1/4 (out of 4 keys, only 1 correct)
P(2nd key is correct) = 1/3 (out of 3 keys left, only 1 correct)
P(3rd key is wrong) = 1/2 (out of 2 keys left, 1 is wrong)
P(4th key is wrong) = 1/1 (1 key left, and it must be wrong because the correct key was put in the 3rd lock)

Since we want all of these events to happen together, we multiply the fractions. We multiply because the more events we want to happen together, the smaller the probability, and when we multiply fractions, the result keeps getting smaller. Thus, letting C = correct and W = wrong:

P(CCWW) = 1/4 * 1/3 * 1/2 * 1/1 = 1/24.

The result above represents the probability of one particular way that 2 keys could be correct: CCWW. Now we need to multiply by the total number of ways that 2 keys could be correct. Any arrangement of the letters CCWW will give us a good outcome:

Number of ways to arrange CCWW = 4!/2!*2! = 6.

Multiplying the results above, we get:

P(exactly 2 keys correct) = 1/24 * 6 = 1/4.

The correct answer is C.
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