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Tony owns six unique matched pairs of socks. All twelve socks are kept

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Tony owns six unique matched pairs of socks. All twelve socks are kept  [#permalink]

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New post 19 Feb 2017, 05:12
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Tony owns six unique matched pairs of socks. All twelve socks are kept loose and unpaired in a drawer. If Tony pulls socks at random, how many must he pull in order to have better than a 50% chance of getting two socks that match?

A) 3

B) 4

C) 5

D) 6

E) 7
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Re: Tony owns six unique matched pairs of socks. All twelve socks are kept  [#permalink]

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New post 01 Mar 2017, 17:16
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Matruco wrote:
Tony owns six unique matched pairs of socks. All twelve socks are kept loose and unpaired in a drawer. If Tony pulls socks at random, how many must he pull in order to have better than a 50% chance of getting two socks that match?

A) 3

B) 4

C) 5

D) 6

E) 7


The probability getting two socks that match can be rephrased as the probability of getting a pair of matching socks or as the probability of getting at least one pair of matching socks. So we can use the the “at least one” concept in probability, that is:

P(at least one pair of matching socks) = 1 - P(no pairs of matching socks)

The probability of no pairs of matching socks (and hence the probability of at least one pair of matching socks) relies on the number of draws. We can see that as P(no pairs of matching socks) decreases, P(at least one pair of matching socks) increases.

After one draw, P(no pairs of matching socks) = 12/12 = 1, since we only draw one sock (we need at least two socks to have a matching pair).

After two draws, P(no pairs of matching socks) = 12/12 x 10/11 = 10/11 (in which 10/11 is the probability of selecting a sock that won’t match the first draw).

After three draws, P(no pairs of matching socks) = 12/12 x 10/11 x 8/10 = 8/11 (8/10 is the probability of drawing the third sock that won't match either of the first two socks).

After four draws, P(no pairs of matching socks) = 12/12 x 10/11 x 8/10 x 6/9 = 16/33 (6/9 is the probability of drawing the fourth sock that won’t match the first three socks).

At this point, we see that P(no pairs of matching socks) = 16/33 < 1/2, so P(at least one pair of matching socks) will be greater than 1/2 since:

P(at least one pair of matching socks) = 1 - 16/33, which is greater than 50%

Thus, we need to have at least 4 draws to have better than a 50% chance of getting two socks that match.

Answer: B
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Re: Tony owns six unique matched pairs of socks. All twelve socks are kept  [#permalink]

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New post 23 Apr 2017, 02:50
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How many socks must Tony pull to guarantee that he will pull at least one pair of socks? The answer is 7 (the worst case scenario - 6 different socks + 1 that has to match one of these 6).

Thus, 7 socks is our 100% probability.
Now, 50% would be simply 7/2 = 3,5.
To be more than 50% and integer, the answer should be 4 -> B.
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Re: Tony owns six unique matched pairs of socks. All twelve socks are kept  [#permalink]

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New post 19 Feb 2017, 05:45
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Official Answer from Veritas Prep;

In any probability problem, there are always at least two possible approaches – one or more direct approaches, and the complementary approach. It’s important to consider which approach will be simplest for any given probability question, especially if a first attempt ends up becoming exceedingly complicated and difficult.

This problem is a perfect example of a case in which a direct approach would be difficult. Two socks that match could be the first and second, or the first and third, or the second and third, or the first and fourth, or the second and fourth, or the third and fourth… Performing and properly combining each calculation would be tedious at best.

Luckily, complementary probability works much better here, as it often does in “at least one” problems (Tony wants at least one pair of socks that match). The opposite outcome is getting no matching socks whatsoever. Just don’t forget to subtract your complementary result from 1 at the end!

Pairs probability can calculate the probability of not matching just as well as the probability of matching. We look at the socks one at a time and ask the simple question, “Are we still happy?” at each point. The goal now is to not get a pair.

The probability that Tony’s first sock will not make a pair is 12/12=1 ;the first sock standing alone is never a pair.

The probability that the second sock will again not make a pair is 10/11. Note that the total socks in the drawer decreased by 1, but the good outcomes (the non-pairing socks) decreased by 2: one sock that Tony picked, and one sock that would pair with the sock he picked. The cumulative probability of no pair is 12/12 ∗ 10/11. which is still a far greater than 50% chance of not having a pair (and a far below 50% chance of having a pair).

The probability that the third sock will again not make a pair is 8/10. Again, the total socks decreased by 1 but the good outcomes decreased by 2, since now two of the ten remaining socks will pair with the two Tony is holding. The cumulative probability of no pair is 12/12 ∗ 10/11 ∗ 8/10 = 8/11, which is still more than a 50% chance of not having a pair (and still less than a 50% chance of having a pair).

The probability that the fourth sock will still not make a pair is 6/9. We know the drill at this point; the total socks decreased by 1, but the good outcomes decreased by 2– the sock Tony just pulled and its mate still in the drawer. The cumulative probability of no pair is 12/12 ∗ 10/11 ∗ 8/10 ∗ 6/9 = 48/99, which is finally less than a 50% chance of not having a pair (and thus more than a 50% chance of having a pair). Picking four socks is enough.

This is answer B.
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Re: Tony owns six unique matched pairs of socks. All twelve socks are kept  [#permalink]

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New post 19 Feb 2017, 06:30
Matruco wrote:
Official Answer from Veritas Prep;

In any probability problem, there are always at least two possible approaches – one or more direct approaches, and the complementary approach. It’s important to consider which approach will be simplest for any given probability question, especially if a first attempt ends up becoming exceedingly complicated and difficult.

This problem is a perfect example of a case in which a direct approach would be difficult. Two socks that match could be the first and second, or the first and third, or the second and third, or the first and fourth, or the second and fourth, or the third and fourth… Performing and properly combining each calculation would be tedious at best.

Luckily, complementary probability works much better here, as it often does in “at least one” problems (Tony wants at least one pair of socks that match). The opposite outcome is getting no matching socks whatsoever. Just don’t forget to subtract your complementary result from 1 at the end!

Pairs probability can calculate the probability of not matching just as well as the probability of matching. We look at the socks one at a time and ask the simple question, “Are we still happy?” at each point. The goal now is to not get a pair.

The probability that Tony’s first sock will not make a pair is 12/12=1 ;the first sock standing alone is never a pair.

The probability that the second sock will again not make a pair is 10/11. Note that the total socks in the drawer decreased by 1, but the good outcomes (the non-pairing socks) decreased by 2: one sock that Tony picked, and one sock that would pair with the sock he picked. The cumulative probability of no pair is 12/12 ∗ 10/11. which is still a far greater than 50% chance of not having a pair (and a far below 50% chance of having a pair).

The probability that the third sock will again not make a pair is 8/10. Again, the total socks decreased by 1 but the good outcomes decreased by 2, since now two of the ten remaining socks will pair with the two Tony is holding. The cumulative probability of no pair is 12/12 ∗ 10/11 ∗ 8/10 = 8/11, which is still more than a 50% chance of not having a pair (and still less than a 50% chance of having a pair).

The probability that the fourth sock will still not make a pair is 6/9. We know the drill at this point; the total socks decreased by 1, but the good outcomes decreased by 2– the sock Tony just pulled and its mate still in the drawer. The cumulative probability of no pair is 12/12 ∗ 10/11 ∗ 8/10 ∗ 6/9 = 48/99, which is finally less than a 50% chance of not having a pair (and thus more than a 50% chance of having a pair). Picking four socks is enough.

This is answer B.



Got it thanks!! Makes sense!!

I highly doubt its easy to think of all this and actually do the math within 2 minutes. Unless you're practicing every possible method of GMAT question over and over again, so that by looking the math, the method just floats in front of your eyes.
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Re: Tony owns six unique matched pairs of socks. All twelve socks are kept  [#permalink]

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New post 27 Feb 2017, 03:46
Matruco wrote:
Tony owns six unique matched pairs of socks. All twelve socks are kept loose and unpaired in a drawer. If Tony pulls socks at random, how many must he pull in order to have better than a 50% chance of getting two socks that match?

A) 3

B) 4

C) 5

D) 6

E) 7



I also faced the same question in veritasprep test, and didnt understood the explaination of veritas prep can someone help please. Have GMAT in 7 Days.
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Re: Tony owns six unique matched pairs of socks. All twelve socks are kept  [#permalink]

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New post 27 Feb 2017, 15:02
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vabzgupta237 wrote:
Matruco wrote:
Tony owns six unique matched pairs of socks. All twelve socks are kept loose and unpaired in a drawer. If Tony pulls socks at random, how many must he pull in order to have better than a 50% chance of getting two socks that match?

A) 3

B) 4

C) 5

D) 6

E) 7



I also faced the same question in veritasprep test, and didnt understood the explaination of veritas prep can someone help please. Have GMAT in 7 Days.


Hey guys. How can I help? What part of the solution are you having trouble with?
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Tony owns six unique matched pairs of socks. All twelve socks are kept  [#permalink]

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New post 27 Feb 2017, 18:24
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vabzgupta237 wrote:
Matruco wrote:
Tony owns six unique matched pairs of socks. All twelve socks are kept loose and unpaired in a drawer. If Tony pulls socks at random, how many must he pull in order to have better than a 50% chance of getting two socks that match?

A) 3

B) 4

C) 5

D) 6

E) 7



I also faced the same question in veritasprep test, and didnt understood the explaination of veritas prep can someone help please. Have GMAT in 7 Days.

EDIT: Sorry vabzgupta237 for some bad math.
Hey vabzgupta237, this is how I thought about it...

On the first pick, you clearly won't have a match, so lets just start at pick #2.

Picked Socks | Drawer Socks

#2 - S | 11S - out of those 11 socks, one is right, so 1/11th chance to be good after 2 picks.
#3 - 2S | 10S - now we have 2 socks on the left, so out of the 10 socks left, 2 will be good - 2/10 or 1/5 , but we have to multiply by the chance of getting to his point, 10/11 so 10/55
#4 - 3S | 9S - out of the 9 socks left, 3 are good picks, so 3/9 or 1/3rd , but we have to multiply by the chance of getting to here, so 1/3 * 10/11 * 8/10 or 8/33

Now, you've got 1/11 + 10/55 + 8/33. Now, lets put these in terms of 11ths. 1/11 + 2/11 + 2.66/11. To be over half, we need the top to equal over 5.5, and its 5 and 2/3rd, so we've got enough.

Therefore, after that 4th pick, you've now got more that 50% chance of being finished.
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Re: Tony owns six unique matched pairs of socks. All twelve socks are kept  [#permalink]

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New post 27 Feb 2017, 20:05
AnthonyRitz wrote:
vabzgupta237 wrote:
Matruco wrote:
Tony owns six unique matched pairs of socks. All twelve socks are kept loose and unpaired in a drawer. If Tony pulls socks at random, how many must he pull in order to have better than a 50% chance of getting two socks that match?

A) 3

B) 4

C) 5

D) 6

E) 7



I also faced the same question in veritasprep test, and didnt understood the explaination of veritas prep can someone help please. Have GMAT in 7 Days.


Hey guys. How can I help? What part of the solution are you having trouble with?


The whole part, how to start following the basic knowledge, not a specialised approach for this question.
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Re: Tony owns six unique matched pairs of socks. All twelve socks are kept  [#permalink]

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New post 27 Feb 2017, 20:24
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vabzgupta237 wrote:
AnthonyRitz wrote:
vabzgupta237 wrote:

I also faced the same question in veritasprep test, and didnt understood the explaination of veritas prep can someone help please. Have GMAT in 7 Days.


Hey guys. How can I help? What part of the solution are you having trouble with?


The whole part, how to start following the basic knowledge, not a specialised approach for this question.


Okay, let me try to walk you through it.

First of all, whenever you're doing a probability problem, you always have at least two possible approaches. One is the direct approach -- calculate the good outcomes. The other is the complementary approach -- calculate the bad outcomes, and then subtract from 1. Either approach will work here, but in cases involving "at least one" (at least one pair of matching socks, etc.) it is often best to go with the complementary approach. The opposite of "at least one" is "none,"and "none" is usually less difficult to calculate than all the different ways of getting "at least one."

Okay, so two approaches. Bradley did a solid job of laying out a direct approach. Pick socks one at a time.

Is the first one a pair? Definitely not.

Is the second one a pair? Only if it matches the first one -- that's one good option out of eleven total, 1/11.

Is the third one a pair? To make it this far, we already are in the case of no pair yet, which is 10/11, and then we are holding two mismatched socks. We will pair one of them 2/10 of the time, so our chance of a good outcome is now 1/11 + 10/11 * 2/10 = 3/11. Note that to get this expression I have applied the Basic Counting Principle -- "and" means "multiply"; "or" means "add." This rule is critical for any problem involving counting. That means almost all combinatorics and much of probability needs this rule.

Is the fourth one a pair? To make it this far, we still have no pair, which happens 8/11 of the time. Of the nine remaining socks, three will match something we have, so the probability is 3/11 + 8/11 * 3/9 = 27/99 + 24/99 = 51/99, which is more than 50%. So four socks is the answer.


The complementary approach focuses on not making a pair.

The first sock avoids a pair 100% of the time.

The next sock avoids a pair 10/11 of the time (everything but the mate of the first sock pulled is fine).

The next sock avoids a pair 8/10 of the time (ten socks are left, but two of them pair the two already pulled). Overall, we have no pair 100% * 10/11 * 8/10 = 8/11 of the time. So we have a pair 1 - 8/11 = 3/11 of the time (note: same result we got the other way).

The next sock avoids a pair 6/9 of the time (note that every step there is one less sock in the drawer but two less *good* socks in the drawer -- not only the one we picked but also its mate will now fail to avoid a pair). Overall, we have no pair 100% * 10/11 * 8/10 * 6/9 = 48/99 of the time. So we have a pair 1 - 48/99 = 51/99 of the time (again, same result with either approach). Four socks is still the right answer.


I want to be clear that these are not fringe or specialized approaches. Complementary probability is a core, fundamental tool for GMAT probability questions. At the 700 level (where this question is squarely located), you often won't have much chance without it. In fact, this problem follows a long line of very similar problems that use these approaches. With practice, they can be done correctly in two minutes or less.

I'm happy to explain further and answer any questions you may have.

I hope this helps!!
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Re: Tony owns six unique matched pairs of socks. All twelve socks are kept  [#permalink]

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New post 27 Feb 2017, 21:35
let me know what wrong in my approach
we AA BB CC DD EE FF ... six pairs
at first suppose picked on is A , now to pick paired one prob is 1/11 , which is far lesser 1/2 =.5
we have to calculate minimum number ..

so another is lets have B ... now any one from (A or B )suffice the need , so prob will 2/10 still less

another one is C , now prob of paired will from (A or B or C) from remaining , will be 3/9 still less

Another one D , now (A or B or C or D)prob will be 4/8 , which is 1/2

Now my confusion is we have to calculate better than 59% that 1/2 ... so i think we have include next ... or shall i consider next will be better , so minimum is 4
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Tony owns six unique matched pairs of socks. All twelve socks are kept  [#permalink]

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New post 27 Feb 2017, 21:51
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I see essentially two problems with this approach.

First of all, somewhere along the line you get a bit confused and end up off by one sock. When you note the 1/11 probability of matching, that's *after* the first sock, so it's really the probability for the second sock. And so forth. The 4/8 probability you cite isn't the probability of the fourth sock matching -- it's the probability of the fifth sock matching (the denominator is 8 because 4 had already been pulled previously). So that's one issue. And yes, if you were counting that way, you'd actually have to go on to the sixth sock to get *better* than 50%. But that brings me to my next issue.

The other, more fundamental issue is that you aren't treating the probability as cumulative. When you say that there's a 4/8 chance of that last sock matching, that's sort of true..... IF you get there at all. It ignores the probability that previous socks might already have matched. The probability of at least one pair among, for instance, four socks isn't just the probability that the fourth sock matches the first, or the fourth matches the second, or the fourth matches the third. It's also the probability that the first and second or first and third or second and third match. You need to be adding all of these probabilities you're getting as you go along. Since you're not, you're consistently underestimating the probability of a pair.
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Re: Tony owns six unique matched pairs of socks. All twelve socks are kept  [#permalink]

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New post 24 Mar 2017, 11:45
I found this question in veritas Prep.
Though one, i admit. I got 5 socks by eliminating socks one by one.

I went through the solution posted by veritas, but iam still looking for an easiest way.
Can anyone help?
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Re: Tony owns six unique matched pairs of socks. All twelve socks are kept  [#permalink]

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New post 18 Apr 2017, 12:54
My approach was to find the probability for each pick for NOT picking a pair. Once this probability drops below 50%, we have our answer:

Pick: 1 2 3 4
Probability of not picking a pair: 1 10/11 8/10 6/9(2/3)
Cumulative probability 1 10/11 8/11 16/33

So, on the 4th pick the probability of not picking a pair drops below 50% and the probability of picking a pair is more than 50%.
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Re: Tony owns six unique matched pairs of socks. All twelve socks are kept  [#permalink]

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New post 09 Jul 2017, 06:04
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Matruco wrote:
Tony owns six unique matched pairs of socks. All twelve socks are kept loose and unpaired in a drawer. If Tony pulls socks at random, how many must he pull in order to have better than a 50% chance of getting two socks that match?

A) 3

B) 4

C) 5

D) 6

E) 7


Probability of getting selected for each socks = \(1/6\) = 16%

In order to find the probability of > 50% lets take each socs and calculate the probability

First Socs = 16%

Second Socs = 32%

Third Socs = 48% (Even if we take exact value of 1/6 = 0.166 or 16.6% - we will still get % as 49.8% which is less than 50%)

Forth Socs = 64%

As we see for fourth socs the probability is more than 50% and hence answer is 4

Hence, Answer is B
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Re: Tony owns six unique matched pairs of socks. All twelve socks are kept  [#permalink]

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New post 11 Feb 2018, 18:18
if we solve this using the veritas prep method, i feel like this would take a REALLY long time. unless you recognize doing this question before and can set it up on the fly. am i wrong?
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Re: Tony owns six unique matched pairs of socks. All twelve socks are kept &nbs [#permalink] 11 Feb 2018, 18:18
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Tony owns six unique matched pairs of socks. All twelve socks are kept

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