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

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A drawer contains 8 pairs of socks. For each sock, there is [#permalink] New post 26 May 2017, 11:13
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Question Stats:

49% (03:42) correct 49% (00:42) wrong based on 14 sessions
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
[Reveal] Spoiler: OA

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4 KUDOS received
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Re: A drawer contains 8 pairs of socks. For each sock, there is [#permalink] New post 29 May 2017, 16:41
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GreenlightTestPrep 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


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(at least 1 pair) = 1 - P(zero pairs)
= 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 [#permalink] New post 21 Dec 2017, 05:45
Can we do as follows
We have 8 pairs so the probability of chosing one pair out of 8 8C1 = 8
Then since we have 16 socks and we are taking 6 socks out of 16. 16C6
The proba will be 8 / 16C6
Can we do with combinatorics?
Because I cant find the result
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Re: A drawer contains 8 pairs of socks. For each sock, there is [#permalink] New post 20 Feb 2019, 08:44
GreenlightTestPrep wrote:


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)


Cheers,
Brent


To understand the difference, could anybody please provide solution

without

using the

complement

rule?
1 KUDOS received
GRE Instructor
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Joined: 10 Apr 2015
Posts: 1444
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Kudos [?]: 1368 [1] , given: 8

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Re: A drawer contains 8 pairs of socks. For each sock, there is [#permalink] New post 20 Feb 2019, 10:04
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Expert's post
AE wrote:
GreenlightTestPrep wrote:


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)


Cheers,
Brent


To understand the difference, could anybody please provide solution

without

using the

complement

rule?


That's a lot of work, but I'll get the ball rolling, so someone else can finish the question.

P(at least 1 pair among the 6 selected socks) = P(1 pair and the other 4 socks are different OR 2 pair and the other 2 socks are different OR 3 pair)
= P(1 pair and the other 4 socks are different) + P(2 pair and the other 2 socks are different) + P(3 pair)

You'll see that things get VERY complicated VERY quickly.

Cheers,
Brent
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Re: A drawer contains 8 pairs of socks. For each sock, there is   [#permalink] 20 Feb 2019, 10:04
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