Problem Solving

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 and without replacement, how many must he pull in order to have better than a 50%
chance of having two socks that match?

3
4
5
6
7


I'm unable to understand how will you know the matched pair of socks. If there were any colour combination it was easier to find as i have searched in other posts.

Kindly guide how can we solve with complementary approach.

Cheers!!!

Kaustubhk 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 and without replacement, how many must he pull in order to have better than a 50% chance of having two socks that match?

3
4
5
6
7

We can PLUG IN THE ANSWERS, which represent the minimum number of socks that must be pulled.
When the correct answer choice is plugged in, the probability of NOT picking a matching pair will be LESS than 1/2 (implying that the probability of picking a matching pair will be MORE than 1/2).
Since we need to determine the minimum number of socks that must be pulled, we should start with the SMALLEST answer choice.

Note:
The first sock pulled can be ANY of the 12 socks and thus is irrelevant.
Our only concern is whether any of the SUBSEQUENT socks form a matching pair.

A: 3 socks pulled
After the 1st sock is pulled:
P(2nd sock does not match the 1st) = 10/11. (Of the 11 socks left, 10 do not match the 1st sock.)
P(3rd sock does not match the 1st or 2nd) = 8/10. (Of the 10 socks left, 8 do not match the 1st or 2nd.)
To combine these probabilities, we multiply:
10/11 * 8/10 = 8/11.
Since the resulting probability is not less than 1/2, eliminate A.

B: 4 socks pulled
After the 1st sock is pulled:
P(2nd sock does not match the 1st) = 10/11. (Of the 11 socks left, 10 do not match the 1st sock.)
P(3rd sock does not match the 1st or 2nd) = 8/10. (Of the 10 socks left, 8 do not match the 1st or 2nd.)
P(4th sock does not match 1st, 2nd, or 3rd) = 6/9. (Of the 9 socks left, 6 do not match the 1st, 2nd or 3rd.)
To combine these probabilities, we multiply:
10/11 * 8/10 * 6/9 = 16/33.
Success!
The resulting probability is less than 1/2.

The correct answer is B.

The OA implies the following:
P(not matching set) = 16/33.
P(matching set) = 1 - 16/33 = 17/33.
The probability in blue is greater than 50%.

Kaustubhk 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 and without replacement, how many must he pull in order to have better than a 50%
chance of having two socks that match?

3
4
5
6
7


I'm unable to understand how will you know the matched pair of socks. If there were any colour combination it was easier to find as i have searched in other posts.

Kindly guide how can we solve with complementary approach.

Cheers!!!

The solution presented above is correct, but I wonder whether it really answers your question. You asked "how will you know the matched pair of socks?"

For ease of understanding, let's assume that the socks are not different colors but rather have a letter on them. We have two socks marked A, two marked B, two marked C, etc., out to F.

So the socks are:

AABBCCDDEEFF (12 total).

You pull out one sock. Let's say that you have pulled out C. Obviously you cannot get a pair by pulling out only one sock! You will have to pull another. What's the chance that this new sock is a C? Well, since there is only 1 C in the drawer and 10 that are not C, your chance is 1/11. Another way of saying this is to say that you have 10/11 chances NOT to pull the right sock.

Let's say that your second pull is a B. You don't have a pair, so you will need to pull another sock. You have two potential pairs. You could pull another B or another C. Either pull would give you a pair. Accordingly you now have 2/10 of pulling a useful sock. Or you can figure that you have an 8/10 chance of pulling a sock that doesn't help you out.

We can figure out the chance of you not pulling a pair by multiplying 10/11 by 8/10. This method has an advantage in that the 10s cancel leaving you with 8/11 chances of not pulling the sock that you need and thus 3/11 chances of pulling the sock that you do need.

Let's imagine that on your first 3 pulls you got an A, a B, and a C. You now have 9 socks left in the drawer, 3 of which will help you and 6 of which will not. If we continue along the theme of calculating the chances that you will not get the sock you need, we can figure that the new sock will give us 6/9 or basically 2/3 chance of being unhelpful.

Since our former probability was 8/11 of not finding a pair, the new probability will be 16/33 of not finding a pair i.e. 17/33 chance of finding a pair.

Hi Mitch,

Thanks for breaking this down for me!!!

1. The note is extremely important because we can only start forming a pair once we remove the second sock out of the drawer.

2. The probability that 2nd Sock doesn't match the 1st also means that 1st Sock has a pair back in the drawer which also means that the other socks(Including the second we removed) do not match. . This took lot of time to understand.

3. I need to solve more of the probability- pairs questions so that it can be aced in the actual GMAT within 2 minutes.


Hi Elias,

As you also rightly pointed out my 2nd point, this is the core of the solution.

Thanks to you both.Cheers!!!

Kaustubhk wrote:I'm unable to understand how will you know the matched pair of socks.

The key is given in the prompt: the pairs are unique. Since the pairs are unique and Tony owns them, he'd presuambly be able to recognize a match when we saw two identical socks in front of him.

It's definitely good to mind details on the GMAT, but don't drive yourself too crazy with the epistemology of socks.

We could also approach this question algebraically.

Let's say that Tony has 2n socks, n unique and distinguishable pairs.

For the first sock, Tony can pull anything.

For the second sock, he must get something that doesn't match the first one, so (2n - 2)/(2n - 1), since he's got (2n - 1) socks left and (2n - 2), or all but one sock, that don't match the first pick.

For the third sock, he must get something that doesn't match either of the first two, so (2n - 4)/(2n - 2), since he's got (2n - 2) socks left and (2n - 4), or all but two socks, that don't match either of the first two.

Notice that this pattern will continue: the second sock gives (2n - 2)/(2n - 1), the third sock gives (2n - 4)/(2n - 2), the fourth sock gives (2n - 6)/(2n - 3), and so on, with the xth sock giving (2n - 2*(x - 1))/(2n - (x - 1)).

With that in mind, if Tony pulls x socks from the drawer, where x ≤ n (if x > n, then Tony has pulled more than half of the socks and must have a match), the probability that he has a match is

1 - ((2n - 2)/(2n - 1) * (2n - 4)/(2n - 2) * (2n - 6)/(2n - 3) * ... * (2n - 2(x - 1))/(2n - (x - 1)))

If we know n and our target probability, we can plug those in. We've got n = 6 and a target > .5, which give us

1 - ((12 - 2)/(12 - 1) * (12 - 4)/(12 - 2) * ... * (12 - 2(x - 1))/(12 - (x - 1)) > .5

and we can solve from there.

Jeff@TargetTestPrep wrote:

Kaustubhk 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 and without replacement, how many must he pull in order to have better than a 50% chance of having two socks that match?

A) 3
B) 4
C) 5
D) 6
E) 7

Determining the number of socks that Tony must pull in order to have a better than 50% chance of having two socks that match is the same as determining the number of socks he must pull in order to have a less than 50% chance that these socks are unmatched. Let's calculate the latter.

For the first sock he pulls, the probability that this sock is unmatched to any other is 1.

For the second sock he pulls, the probability that this sock is unmatched to the first one is 10/11 (since there are 11 socks left after the first sock and 10 do not match the first sock). Thus, the probability that the two socks are unmatched is 1 x 10/11 = 10/11.

For the third sock he pulls, the probability that this sock does not match either of the first two is 8/10 (since there are 10 socks left after the first two socks and 8 of them do not match). Thus, the probability that the three socks do not match is 1 x 10/11 x 8/10 = 8/11.

For the fourth sock he pulls, the probability that this sock does not match the first three is 6/9 (since there are 9 socks left after the first three socks and 6 of them do not match). Thus, the probability that the four socks do not match is 1 x 10/11 x 8/10 x 6/9 = 48/99, which is less than 48/96 or 0.5.

Thus, we see that if he pulls 4 socks, the probability that these socks do not match is less than 50%. In other words, the probability that two of them will match must be more than 50%.

Answer: B

Hi Jeff,

Can we approach the question as below

If we select 7 stocks (6+1) then there will surely be a pair and hence 100% probability(since of 12 socks,only 6 are unique).So prbab that its greater than 50% is 7/2 =3.5 Now a nearest integer unit greater than 3.5 is 4.So,we need make a minimum selection of 4 units of single socks.Hence OA is B