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5 pairs of socks are randomly distributed in a bag

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

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New post 25 Aug 2018, 22:52
1
5
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A
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D
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Difficulty:

  25% (medium)

Question Stats:

67% (00:41) correct 33% (00:47) wrong based on 101 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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Re: 5 pairs of socks are randomly distributed in a bag  [#permalink]

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New post 25 Aug 2018, 23:36
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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Re: 5 pairs of socks are randomly distributed in a bag  [#permalink]

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New post 27 Aug 2018, 20:05
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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Re: 5 pairs of socks are randomly distributed in a bag  [#permalink]

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New post 27 Aug 2018, 21:20
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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PKN

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

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New post 27 Aug 2018, 23:54
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
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5 pairs of socks are randomly distributed in a bag  [#permalink]

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New post 28 Aug 2018, 00:22
1
1
1
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
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5 pairs of socks are randomly distributed in a bag  [#permalink]

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New post 28 Aug 2018, 00:26
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 --- > \(\frac{10!}{2!(10-2)!} = 45\)

\(\frac{5}{45}\) = \(\frac{1}{9}\)
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Re: 5 pairs of socks are randomly distributed in a bag  [#permalink]

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New post 10 Sep 2018, 05:52
Hi stne

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

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New post 10 Sep 2018, 06:00
CounterSniper wrote:
Hi stne

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

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New post 10 Sep 2018, 06:05
1
stne wrote:
CounterSniper wrote:
Hi stne

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

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New post 10 Sep 2018, 06:07
CounterSniper wrote:
stne wrote:
CounterSniper wrote:
Hi stne

Please find below
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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Re: 5 pairs of socks are randomly distributed in a bag &nbs [#permalink] 10 Sep 2018, 06:07
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5 pairs of socks are randomly distributed in a bag

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