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# A drawer contains 8 pairs of socks. For each sock, there is exactly

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CEO
Joined: 12 Sep 2015
Posts: 3848
A drawer contains 8 pairs of socks. For each sock, there is exactly  [#permalink]

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26 May 2017, 12:12
5
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31
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Difficulty:

85% (hard)

Question Stats:

55% (03:05) correct 45% (02:50) wrong based on 162 sessions

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A drawer contains 8 pairs of socks. For each sock, there is exactly one matching sock. If Ed randomly selects 6 socks without replacement, what is the probability that he will have at least one pair of matching socks?

A) 1/4
B) 19/51
C) 111/143
D) 3/4
E) 145/189

*kudos for all correct solutions

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Joined: 12 Sep 2015
Posts: 3848
Re: A drawer contains 8 pairs of socks. For each sock, there is exactly  [#permalink]

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27 May 2017, 06:50
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Top Contributor
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GMATPrepNow wrote:
A drawer contains 8 pairs of socks. For each sock, there is exactly one matching sock. If Ed randomly selects 6 socks without replacement, what is the probability that he will have at least one pair of matching socks?

A) 1/4
B) 19/51
C) 111/143
D) 3/4
E) 145/189

*kudos for all correct solutions

Here's an approach that uses probability rules.

When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)

So, here we get: P(at least 1 matching pair) = 1 - P(zero matching pairs)

P(zero matching pairs)
P(zero matching pairs) = P(pick ANY sock 1st AND 2nd sock doesn't match 1st sock AND 3rd sock doesn't match other selections AND 4th sock doesn't match other selections AND 5th sock doesn't match other selections AND 6th sock doesn't match other selections)
= P(pick ANY sock 1st) x P(2nd sock doesn't match 1st sock) x P(3rd sock doesn't match other selections) x P(4th sock doesn't match other selections) x P(5th sock doesn't match other selections) x P(6th sock doesn't match other selections)
= 1 x 14/15 x 12/14 x 10/13 x 8/12 x 6/11
= 32/143

So, P(win at least 1 prize) = 1 - P(win zero prizes)
= 1 - 32/143
= 111/143

ASIDE: Once we draw the first sock, there are 15 sock s remaining, and only 1 matches the 1st selection.
This means there are 14 sock s that DON'T match the first.
So, P(no match on 2nd draw) = 14/15

Then, once we draw the second sock (without a match), there are 14 socks remaining, and 2 of them match either the 1st or 2nd selection.
This means there are 12 socks that DON'T match.
So, P(no match on 3rd draw) = 12/14

etc

Cheers,
Brent
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A drawer contains 8 pairs of socks. For each sock, there is exactly  [#permalink]

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13 Nov 2017, 05:09
5
Probability = 1 - P(no pair)

$$P(no pair) = \frac{Ways To Select 6 Unique Socks}{Ways To Select Any Six}$$

$$P(no pair) = \frac{16 *14*12*10*8*6}{16*15*14*13*12*11}$$

$$P(no pair) = \frac{32}{143}$$

Probability = 1 - 32/143
= 111/143

C.
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26 May 2017, 12:33
2
Prob of choosing 6 socks so that there is no pair among them= 8C6 (2C1)^6
We are doing 2c1 coz any one one sock can b chosen among a pair
P (atleast one pair )= 1- p (no pair)

= 1- [ 8c6 * 2^6] / 16c6
=111/143

Posted from my mobile device
Intern
Joined: 14 Sep 2016
Posts: 16
Re: A drawer contains 8 pairs of socks. For each sock, there is exactly  [#permalink]

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13 Nov 2017, 04:28
GMATPrepNow wrote:
GMATPrepNow wrote:
A drawer contains 8 pairs of socks. For each sock, there is exactly one matching sock. If Ed randomly selects 6 socks without replacement, what is the probability that he will have at least one pair of matching socks?

A) 1/4
B) 19/51
C) 111/143
D) 3/4
E) 145/189

*kudos for all correct solutions

Here's an approach that uses probability rules.

When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)

So, here we get: P(at least 1 matching pair) = 1 - P(zero matching pairs)

P(zero matching pairs)
P(zero matching pairs) = P(pick ANY sock 1st AND 2nd sock doesn't match 1st sock AND 3rd sock doesn't match other selections AND 4th sock doesn't match other selections AND 5th sock doesn't match other selections AND 6th sock doesn't match other selections)
= P(pick ANY sock 1st) x P(2nd sock doesn't match 1st sock) x P(3rd sock doesn't match other selections) x P(4th sock doesn't match other selections) x P(5th sock doesn't match other selections) x P(6th sock doesn't match other selections)
= 1 x 14/15 x 12/14 x 10/13 x 8/12 x 6/11
= 32/143

So, P(win at least 1 prize) = 1 - P(win zero prizes)
= 1 - 32/143
= 111/143

ASIDE: Once we draw the first sock, there are 15 sock s remaining, and only 1 matches the 1st selection.
This means there are 14 sock s that DON'T match the first.
So, P(no match on 2nd draw) = 14/15

Then, once we draw the second sock (without a match), there are 14 socks remaining, and 2 of them match either the 1st or 2nd selection.
This means there are 12 socks that DON'T match.
So, P(no match on 3rd draw) = 12/14

etc

Cheers,
Brent

Hi Brent,

I tried to use combinations. What am I doing wrong?

P(at least 1 pair) = 1 - P(no pairs)

P(no pairs) = (No. of ways 6 unpaired socks can be picked from 8 unpaired socks)/(total number of ways 6 socks can be picked from 16 socks)

= 8C6/16C6
= 1/286

P(at least 1 pair) = 1 - 1/286 = 285/286

Of course, answer not among choices.
CEO
Joined: 12 Sep 2015
Posts: 3848
Re: A drawer contains 8 pairs of socks. For each sock, there is exactly  [#permalink]

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13 Nov 2017, 08:48
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Top Contributor
1
dramamur wrote:
Hi Brent,

I tried to use combinations. What am I doing wrong?

P(at least 1 pair) = 1 - P(no pairs)

P(no pairs) = (No. of ways 6 unpaired socks can be picked from 8 unpaired socks)/(total number of ways 6 socks can be picked from 16 socks)

= 8C6/16C6
= 1/286

P(at least 1 pair) = 1 - 1/286 = 285/286

Of course, answer not among choices.

Hi dramamur,

The problem with your solution is that you are treating the socks differently when calculating the numerator and denominator.
For the denominator, you are treating the 16 socks as being 100% unique. It's as though you are saying that each pair of matching socks has a LEFT sock and a RIGHT sock.
So, we can select 6 socks from 16 UNIQUE socks in 16C6 ways.

However, for the numerator, you are essentially saying that the two socks in each matching pair are identical.
You are not making a LEFT/RIGHT distinction for each pair. Instead, you are choosing 6 COLORS from a total of 8 COLORS.

Let's keep going with your solution, BUT we'll make sure we consider the fact that we have a LEFT and RIGHT sock for each pair.

So, we'll have to alter your numerator somewhat.
First, we can select 6 COLORS from a total of 8 COLORS in 6C2 ways (28 ways)
For the 1st COLOR, we can choose either the LEFT sock or the RIGHT sock. We can do this in 2 ways.
For the 2nd COLOR, we can choose either the LEFT sock or the RIGHT sock. We can do this in 2 ways.
For the 3rd COLOR, we can choose either the LEFT sock or the RIGHT sock. We can do this in 2 ways.
For the 4th COLOR, we can choose either the LEFT sock or the RIGHT sock. We can do this in 2 ways.
For the 5th COLOR, we can choose either the LEFT sock or the RIGHT sock. We can do this in 2 ways.
For the 6th COLOR, we can choose either the LEFT sock or the RIGHT sock. We can do this in 2 ways.
So, the total number of ways to select 6 UNMATCHED socks = (28)(2)(2)(2)(2)(2)(2)

We'll keep the numerator the same.
That is, we can select 6 socks from 16 UNIQUE socks in 16C6 ways.

So, P(ZERO matching socks) = (28)(2)(2)(2)(2)(2)(2) /16C6
= 32/143

P(at least 1 pair) = 1 - P(no pairs)
= 1 - 32/143
= 111/143

Cheers,
Brent
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Posts: 8
A drawer contains 8 pairs of socks. For each sock, there is exactly  [#permalink]

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03 Jun 2019, 14:45
I don't get why this is wrong:

probability he won't pick a matching sock:

(8/8 * 6/7 *5/6*4/5*3/4*2/3) = 2/7 that he won't pick a matching sock.

I guess i didn't factor in the fact that we have L and R socks. I assumed they were of different colors.

in that case it would be (16/16 * 14/15 * 12/14 * 10/13 * 8/12 *6/11)= 32/143

1- 32/143= 111/43
CEO
Joined: 12 Sep 2015
Posts: 3848
A drawer contains 8 pairs of socks. For each sock, there is exactly  [#permalink]

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Updated on: 04 Jun 2019, 05:20
Top Contributor
mgmat4944 wrote:
I don't get why this is wrong:

probability he won't pick a matching sock:

(8/8 * 6/7 *5/6*4/5*3/4*2/3) = 2/7 that he won't pick a matching sock.

I guess i didn't factor in the fact that we have L and R socks. I assumed they were of different colors.

in that case it would be (16/16 * 14/15 * 12/14 * 10/13 * 8/12 *6/11)= 32/143

1- 32/143= 111/43

The problem starts with the part in blue above

P(Choose any sock 1st) 16/16 = 1
P(Choose 2nd sock that does NOT match 1st sock) = 14/15
P(Choose 3rd sock that does NOT match 1st sock OR the 2nd sock) = 12/14 (there are 14 socks remaining, and 1 sock matches 1st sock, and 1 sock matches 2nd sock)
P(Choose 4th sock that does NOT match 1st sock OR the 2nd sock OR the 3rd sock) = 10/13 (there are 13 socks remaining, and 1 sock matches 1st sock, and 1 sock matches 2nd sock, and 1 sock matches 3rd sock)
etc

Cheers,
Brent
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Originally posted by GMATPrepNow on 03 Jun 2019, 15:49.
Last edited by GMATPrepNow on 04 Jun 2019, 05:20, edited 1 time in total.
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Re: A drawer contains 8 pairs of socks. For each sock, there is exactly  [#permalink]

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04 Jun 2019, 00:42
GMATPrepNow wrote:
A drawer contains 8 pairs of socks. For each sock, there is exactly one matching sock. If Ed randomly selects 6 socks without replacement, what is the probability that he will have at least one pair of matching socks?

A) 1/4
B) 19/51
C) 111/143
D) 3/4
E) 145/189

*kudos for all correct solutions

8 pairs = 16 socks
1- 16/16*14/15*12/14*10/13*8/12*6/11 = 111/143
IMO C
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Re: A drawer contains 8 pairs of socks. For each sock, there is exactly  [#permalink]

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25 Jun 2019, 14:12
Hi ,

Can you tell me whats wrong with my solution
Chances of picking the 1 st sock =16
Chances of picking the 2 nd sock =14

and so on

So my final answer was 1-(16*14*12*10*8*6/16C6)
What's wrong with my approach?
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Re: A drawer contains 8 pairs of socks. For each sock, there is exactly  [#permalink]

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25 Jun 2019, 16:45
1
Because order of the socks doesn't matter, you have to divide (16*14*12*10*8*6) by 6!.

Ambika02 wrote:
Hi ,

Can you tell me whats wrong with my solution
Chances of picking the 1 st sock =16
Chances of picking the 2 nd sock =14

and so on

So my final answer was 1-(16*14*12*10*8*6/16C6)
What's wrong with my approach?
Intern
Joined: 14 Jun 2019
Posts: 2
Re: A drawer contains 8 pairs of socks. For each sock, there is exactly  [#permalink]

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14 Jul 2019, 03:11
1
Ambika02 wrote:
Hi ,

Can you tell me whats wrong with my solution
Chances of picking the 1 st sock =16
Chances of picking the 2 nd sock =14

and so on

So my final answer was 1-(16*14*12*10*8*6/16C6)
What's wrong with my approach?

because you are mixing two approaches probability and combination one. chance of picking the first sock is 16/16 and second is 14/15 every sock have different you can't take just 16C6
Re: A drawer contains 8 pairs of socks. For each sock, there is exactly   [#permalink] 14 Jul 2019, 03:11
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