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# 5 pairs of socks are randomly distributed in a bag

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Manager
Joined: 28 Jun 2018
Posts: 153
Location: India
Concentration: Finance, Marketing
Schools: CUHK '21 (II)
GMAT 1: 650 Q49 V30
GPA: 4
5 pairs of socks are randomly distributed in a bag  [#permalink]

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25 Aug 2018, 23:52
2
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Difficulty:

15% (low)

Question Stats:

72% (00:55) correct 28% (01:11) wrong based on 159 sessions

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5 pairs of socks are randomly distributed in a bag.If you reach in and select 2 socks, what is the probability that you are holding a matched pair?
a) 1/100
b) 1/90
c) 1/10
d) 1/9
e) 1/5

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Status: Learning stage
Joined: 01 Oct 2017
Posts: 1019
WE: Supply Chain Management (Energy and Utilities)
Re: 5 pairs of socks are randomly distributed in a bag  [#permalink]

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26 Aug 2018, 00:36
1
2
GMATbuster92 wrote:
5 pairs of socks are randomly distributed in a bag.If you reach in and select 2 socks, what is the probability that you are holding a matched pair?
a) 1/100
b) 1/90
c) 1/10
d) 1/9
e) 1/5

Sample space= No of ways of selecting 2 nos. of socks out of 10 nos. of socks=10C2

Favourable ways= No of ways selecting 1 pair of socks out of 5 pairs=5C1

So, required probability=$$\frac{5C1}{10C2}$$=$$\frac{5}{5*9}$$=$$\frac{1}{9}$$

Ans. (D)
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PKN

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Re: 5 pairs of socks are randomly distributed in a bag  [#permalink]

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27 Aug 2018, 21:05
PKN wrote:
GMATbuster92 wrote:
5 pairs of socks are randomly distributed in a bag.If you reach in and select 2 socks, what is the probability that you are holding a matched pair?
a) 1/100
b) 1/90
c) 1/10
d) 1/9
e) 1/5

Sample space= No of ways of selecting 2 nos. of socks out of 10 nos. of socks=10C2

Favourable ways= No of ways selecting 1 pair of socks out of 5 pairs=5C1

So, required probability=$$\frac{5C1}{10C2}$$=$$\frac{5}{5*9}$$=$$\frac{1}{9}$$

Ans. (D)

Hi,

Are you not comparing apples against the oranges? I mean, blue above says socks and green says pairs?

I used this technique, please correct me if I am wrong.

Base:- 10C2- 5 Pairs-->10 Socks and 2 Selections - Makes Sense
Fav Outcomes: 2C2*5 [1 Option, 5 Pairs]

Prob- 5/45=1/9

"D"

Thanks,
Vikram
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Status: Learning stage
Joined: 01 Oct 2017
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WE: Supply Chain Management (Energy and Utilities)
Re: 5 pairs of socks are randomly distributed in a bag  [#permalink]

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27 Aug 2018, 22:20
1
vikgupta07 wrote:
PKN wrote:
GMATbuster92 wrote:
5 pairs of socks are randomly distributed in a bag.If you reach in and select 2 socks, what is the probability that you are holding a matched pair?
a) 1/100
b) 1/90
c) 1/10
d) 1/9
e) 1/5

Sample space= No of ways of selecting 2 nos. of socks out of 10 nos. of socks=10C2

Favourable ways= No of ways selecting 1 pair of socks out of 5 pairs=5C1

So, required probability=$$\frac{5C1}{10C2}$$=$$\frac{5}{5*9}$$=$$\frac{1}{9}$$

Ans. (D)

Hi,

Are you not comparing apples against the oranges? I mean, blue above says socks and green says pairs?

I used this technique, please correct me if I am wrong.

Base:- 10C2- 5 Pairs-->10 Socks and 2 Selections - Makes Sense
Fav Outcomes: 2C2*5 [1 Option, 5 Pairs]

Prob- 5/45=1/9

"D"

Thanks,
Vikram

Hi vikgupta07 .

aa bb cc dd ee

It's not about $$\frac{#pairs}{#socks}$$ or $$\frac{#socks}{#socks}$$, rather it is $$\frac{#pairs}{#pairs}$$, where
Numerator=favorable events(in pairs)= No of times or ways we can hold one matching pair of socks out of 5 pairs=5C1 (aa,bb,cc,dd,ee)
Denominator=Total no of events(in pairs)=No of times or ways we can hold a pair of socks(selection of 2nos) out of 10 socks=10C2 (The outcomes of 10C2 are in pair. Such as ab, ac, aa,bb,bc,bd etc [Matched pair+Unmatched pair])

Hope its clear.

P.S:- Not sure how you have arrived at 2C2*5.
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Regards,

PKN

Rise above the storm, you will find the sunshine
Intern
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Re: 5 pairs of socks are randomly distributed in a bag  [#permalink]

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28 Aug 2018, 00:54
Thanks for the reply.

For the numerator part, choosing two correct socks out of a pair: 2C2

The choice will be a ‘equally likely’ criterion for all the 5 pairs.

2C2+2C2+2C2+2C2+2C2

And hence 5*2C2.

Still working on my probability concepts. Thanks for bearing with me ?

Posted from my mobile device
Director
Joined: 20 Feb 2015
Posts: 767
Concentration: Strategy, General Management
5 pairs of socks are randomly distributed in a bag  [#permalink]

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28 Aug 2018, 01:22
1
1
2
GMATbuster92 wrote:
5 pairs of socks are randomly distributed in a bag.If you reach in and select 2 socks, what is the probability that you are holding a matched pair?
a) 1/100
b) 1/90
c) 1/10
d) 1/9
e) 1/5

1st sock can be selected in 10 ways
2nd one can be selected in only one way
p(1st sock) = 10/10
p(2nd sock) = 1/9

p(of selecting 2 of same colour) = 10/90 = 1/9
VP
Joined: 09 Mar 2016
Posts: 1230
5 pairs of socks are randomly distributed in a bag  [#permalink]

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28 Aug 2018, 01:26
GMATbuster92 wrote:
5 pairs of socks are randomly distributed in a bag.If you reach in and select 2 socks, what is the probability that you are holding a matched pair?
a) 1/100
b) 1/90
c) 1/10
d) 1/9
e) 1/5

here is my approach guys

We have following pair of socks: AA, BB, CC, DD, EE hence we have FIVE possibilities to choose a a matching pair of sock

Total number of socks 10

Total number of possibilities to choose any two socks from 10 --- > $$\frac{10!}{2!(10-2)!} = 45$$

$$\frac{5}{45}$$ = $$\frac{1}{9}$$
Director
Joined: 20 Feb 2015
Posts: 767
Concentration: Strategy, General Management
Re: 5 pairs of socks are randomly distributed in a bag  [#permalink]

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10 Sep 2018, 06:52
Hi stne

another way similar to the one I have posted

PKN has already given an appropriate explanation , I just wanted to show another way to solve the question

as per the question we have 5 pairs
consider below

11------------22----------------33-------------------44-------------55

are the given pairs

no of ways we can select one pair = 5c1 = 5 ways
total number of ways = 10c2=45

probability = 5/45 = 1/9

I have solved this question the same way as I have done below
https://gmatclub.com/forum/jessica-s-la ... s#p2128841
Director
Joined: 27 May 2012
Posts: 902
Re: 5 pairs of socks are randomly distributed in a bag  [#permalink]

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10 Sep 2018, 07:00
CounterSniper wrote:
Hi stne

another way similar to the one I have posted

PKN has already given an appropriate explanation , I just wanted to show another way to solve the question

as per the question we have 5 pairs
consider below

11------------22----------------33-------------------44-------------55

are the given pairs

no of ways we can select one pair = 5c1 = 5 ways
total number of ways = 10c2=45

probability = 5/45 = 1/9

I have solved this question the same way as I have done below
https://gmatclub.com/forum/jessica-s-la ... s#p2128841

Actually this seems different from the other post.

Ways of selecting the first sock is 10C1 for the second sock there is one one way
hence 10C1 *1
Total ways of selecting 2 socks from 10 socks is 10C2 = 45
hence probability is 10/45 = 2/9

https://gmatclub.com/forum/jessica-s-la ... l#p2128841
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- Stne
Director
Joined: 20 Feb 2015
Posts: 767
Concentration: Strategy, General Management
Re: 5 pairs of socks are randomly distributed in a bag  [#permalink]

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10 Sep 2018, 07:05
1
stne wrote:
CounterSniper wrote:
Hi stne

another way similar to the one I have posted

PKN has already given an appropriate explanation , I just wanted to show another way to solve the question

as per the question we have 5 pairs
consider below

11------------22----------------33-------------------44-------------55

are the given pairs

no of ways we can select one pair = 5c1 = 5 ways
total number of ways = 10c2=45

probability = 5/45 = 1/9

I have solved this question the same way as I have done below
https://gmatclub.com/forum/jessica-s-la ... s#p2128841

Actually this seems different from the other post.

Ways of selecting the first sock is 10C1 for the second sock there is one one way
hence 10C1 *1
Total ways of selecting 2 socks from 10 socks is 10C2 = 45
hence probability is 10/45 = 2/9

https://gmatclub.com/forum/jessica-s-la ... l#p2128841

when we select the first sock , only 9 are left !!
so the remaining one can be selected in 9c1=9 ways
so probability - 1/9
Director
Joined: 27 May 2012
Posts: 902
Re: 5 pairs of socks are randomly distributed in a bag  [#permalink]

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10 Sep 2018, 07:07
CounterSniper wrote:
stne wrote:
CounterSniper wrote:
Hi stne

another way similar to the one I have posted

PKN has already given an appropriate explanation , I just wanted to show another way to solve the question

as per the question we have 5 pairs
consider below

11------------22----------------33-------------------44-------------55

are the given pairs

no of ways we can select one pair = 5c1 = 5 ways
total number of ways = 10c2=45

probability = 5/45 = 1/9

I have solved this question the same way as I have done below
https://gmatclub.com/forum/jessica-s-la ... s#p2128841

Actually this seems different from the other post.

Ways of selecting the first sock is 10C1 for the second sock there is one one way
hence 10C1 *1
Total ways of selecting 2 socks from 10 socks is 10C2 = 45
hence probability is 10/45 = 2/9

https://gmatclub.com/forum/jessica-s-la ... l#p2128841

when we select the first sock , only 9 are left !!
so the remaining one can be selected in 9c1=9 ways
so probability - 1/9

OOps ! My bad , realized my mistake. Thank you.
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Posts: 42
Re: 5 pairs of socks are randomly distributed in a bag  [#permalink]

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22 Dec 2018, 16:12
GMATbuster92 wrote:
5 pairs of socks are randomly distributed in a bag.If you reach in and select 2 socks, what is the probability that you are holding a matched pair?
a) 1/100
b) 1/90
c) 1/10
d) 1/9
e) 1/5

Probability of the first one don't not matter, so we should consider what is the probability of the second one be the pair of the first one. >>1/9
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Re: 5 pairs of socks are randomly distributed in a bag   [#permalink] 22 Dec 2018, 16:12
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