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

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

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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

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[Reveal] Spoiler: OA

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New post 26 May 2017, 11:33
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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

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

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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

Answer:
[Reveal] Spoiler:
C


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

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New post 13 Nov 2017, 03:28
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

Answer:
[Reveal] Spoiler:
C


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. :(

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

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

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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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Kudos [?]: 2773 [1], given: 364

Re: A drawer contains 8 pairs of socks. For each sock, there is exactly   [#permalink] 13 Nov 2017, 07:48
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