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Tony owns six unique matched pairs of socks. All twelve sock

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by BTGmoderatorDC » Thu May 02, 2019 1:07 am

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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

OA B

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by Ian Stewart » Thu May 02, 2019 1:37 am
We can work out the probability he continues to get unmatched socks, and once that probability falls below 1/2, we'll know he has a greater than 1/2 chance of getting at least one pair of matched socks.

The first sock he picks doesn't matter. The next sock has a 10/11 chance of not matching the first. Now he has two different socks, so of the 10 that remain, only 8 do not match the first two selections, so an 8/10 probability of having no matched pair. Then he has three that don't match, and only 6 of the 9 remaining do not match, for a 6/9 probability of no match. If we multiply these probabilities:

(10/11)(8/10)(6/9) = (1/11)(8/1)(2/3) = 16/33

this probability is less than 1/2, so once he picks four socks, the probability is 17/33 that he has at least one pair of matched socks.
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by GMATGuruNY » Thu May 02, 2019 2:03 am
BTGmoderatorDC 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
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%.
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by GMATGuruNY » Thu May 02, 2019 2:03 am
BTGmoderatorDC 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
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%.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

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by Scott@TargetTestPrep » Mon May 06, 2019 7:12 pm
BTGmoderatorDC 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

OA B

Source: Veritas Prep
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

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