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A drawer has six loose blue socks and six loose white socks. [#permalink]

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23 Sep 2012, 08:06

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A drawer has six loose blue socks and six loose white socks. If four socks are removed from the drawer at random and without replacement. What is the probability that one pair of each color was selected?

Re: A drawer has six loose blue socks and six loose white socks. [#permalink]

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23 Sep 2012, 08:59

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Expert's post

solo1234 wrote:

A drawer has six loose blue socks and six loose white socks. If four socks are removed from the drawer at random and without replacement. What is the probability that one pair of each color was selected?

A. 2/33 B. 5/66 C. 5/33 D. 5/11 E. 1/2

So, we want the probability of removing 2 blues socks out of 6 and 2 white socks out of 6, while removing 4 socks out of 12.

Re: A drawer has six loose blue socks and six loose white socks. [#permalink]

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23 Sep 2012, 19:05

Bunuel wrote:

solo1234 wrote:

A drawer has six loose blue socks and six loose white socks. If four socks are removed from the drawer at random and without replacement. What is the probability that one pair of each color was selected?

A. 2/33 B. 5/66 C. 5/33 D. 5/11 E. 1/2

So, we want the probability of removing 2 blues socks out of 6 and 2 white socks out of 6, while removing 4 socks out of 12.

Re: A drawer has six loose blue socks and six loose white socks. [#permalink]

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25 Sep 2012, 00:27

No of Loose blue socks = 6 (3 pairs) No. of Loose white Socks = 6 (3 pairs) total no. of blue and white socks = 12 Total no. of socks to be selected = 4

So we have C (12,4) = no of ways the socks can be selected = 495

No of ways One pair of blue socks is selected ( 2 blue socks) , C (6,2) , No of ways one pair of white socks can be selected C (6,2) ..Because we have to find a scenario where EXACTLY one pair of Blue socks and ONE pair of WHITE socks is selected we will multiply the two .. ie 15 x 15 ..

Filling the information in the Probability formula we get P (A) = (15 x 15) / 495 = 5 : 11 (D) _________________

"When you want to succeed as bad as you want to breathe, then you’ll be successful.” - Eric Thomas

Re: A drawer has six loose blue socks and six loose white socks. [#permalink]

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26 May 2013, 09:43

Bunuel wrote:

solo1234 wrote:

A drawer has six loose blue socks and six loose white socks. If four socks are removed from the drawer at random and without replacement. What is the probability that one pair of each color was selected?

A. 2/33 B. 5/66 C. 5/33 D. 5/11 E. 1/2

So, we want the probability of removing 2 blues socks out of 6 and 2 white socks out of 6, while removing 4 socks out of 12.

Re: A drawer has six loose blue socks and six loose white socks. [#permalink]

Show Tags

27 May 2013, 00:48

Expert's post

apd2006 wrote:

Bunuel wrote:

solo1234 wrote:

A drawer has six loose blue socks and six loose white socks. If four socks are removed from the drawer at random and without replacement. What is the probability that one pair of each color was selected?

A. 2/33 B. 5/66 C. 5/33 D. 5/11 E. 1/2

So, we want the probability of removing 2 blues socks out of 6 and 2 white socks out of 6, while removing 4 socks out of 12.

Re: A drawer has six loose blue socks and six loose white socks. [#permalink]

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27 May 2013, 03:08

But our answers differ. Which approach/answer is correct? In your approach to solution ,you are replacing (taking number of different combinations of 4 at a time, that means you are replacing the socks back, otherwise why is the count of socks not decreasing? and this is done in both -while calculating the favorable and total outcomes.)

Re: A drawer has six loose blue socks and six loose white socks. [#permalink]

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27 May 2013, 03:59

Expert's post

apd2006 wrote:

But our answers differ. Which approach/answer is correct? In your approach to solution ,you are replacing (taking number of different combinations of 4 at a time, that means you are replacing the socks back, otherwise why is the count of socks not decreasing? and this is done in both -while calculating the favorable and total outcomes.)

Your approach is wrong.

\(C^2_6\) is picking 2 out of 6 without replacement.

If the socks are identical, then the number of ways of selecting a sock in any turn would be 2 (either a W or B). If they form a pair they will be identical?

IMO - all possible selections are - {bbbb, wwww, bbbw, wwwb, wwbb} out of which wwbb is what we require ~ hence answer should be 1/5?

For e.g. if white and blue here would be boys and girls and then we are asked to find a team which has exactly 2 boys and 2 girls - In this case, the solution you mentioned should be valid.

Can you please explain.

Thank you

Bunuel wrote:

solo1234 wrote:

A drawer has six loose blue socks and six loose white socks. If four socks are removed from the drawer at random and without replacement. What is the probability that one pair of each color was selected?

A. 2/33 B. 5/66 C. 5/33 D. 5/11 E. 1/2

So, we want the probability of removing 2 blues socks out of 6 and 2 white socks out of 6, while removing 4 socks out of 12.

Re: A drawer has six loose blue socks and six loose white socks. [#permalink]

Show Tags

12 Dec 2015, 07:35

Expert's post

rsaahil90 wrote:

Hello Bunuel

If the socks are identical, then the number of ways of selecting a sock in any turn would be 2 (either a W or B). If they form a pair they will be identical?

IMO - all possible selections are - {bbbb, wwww, bbbw, wwwb, wwbb} out of which wwbb is what we require ~ hence answer should be 1/5?

For e.g. if white and blue here would be boys and girls and then we are asked to find a team which has exactly 2 boys and 2 girls - In this case, the solution you mentioned should be valid.

Can you please explain.

Thank you

Bunuel wrote:

solo1234 wrote:

A drawer has six loose blue socks and six loose white socks. If four socks are removed from the drawer at random and without replacement. What is the probability that one pair of each color was selected?

A. 2/33 B. 5/66 C. 5/33 D. 5/11 E. 1/2

So, we want the probability of removing 2 blues socks out of 6 and 2 white socks out of 6, while removing 4 socks out of 12.

Hi rsaahil90, you may be correct in the 5 types of combination .. however you have 6 pairs from which you have to choose these combinations and each combination does not have same weightage... lets see this question only.. combinations .. 1) bbbb- 6C4.. choosing 4 black socks out of avail 6= 15.. 2)wwww- same as 1)-15 3)bbbw-6C3*6C1=120 4)wwwb-same as 3)=120 5)wwbb-6C2*6C2=15*15=225

now total ways =15+15+120+120+225=495.. the wwbb way=225 so prob=225/495=5/11.. hope the concept was clear.. _________________

Re: A drawer has six loose blue socks and six loose white socks. [#permalink]

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12 Dec 2015, 07:44

Hello Chetan

This is my whole concern - if all six of the socks are identical then there is just one way to choose 2 or 3 or 6 socks from that color. For e.g. if you have 5 red colored balls, in how many ways you can choose three i.e. 1 way as all the red balls are identical. If in this example balls would have been boys, then 5C3 ways are possible but not in case of red balls or socks in the concerned question

Thanks

Bunuel wrote:

solo1234 wrote:

A drawer has six loose blue socks and six loose white socks. If four socks are removed from the drawer at random and without replacement. What is the probability that one pair of each color was selected?

A. 2/33 B. 5/66 C. 5/33 D. 5/11 E. 1/2

So, we want the probability of removing 2 blues socks out of 6 and 2 white socks out of 6, while removing 4 socks out of 12.

Hi rsaahil90, you may be correct in the 5 types of combination .. however you have 6 pairs from which you have to choose these combinations and each combination does not have same weightage... lets see this question only.. combinations .. 1) bbbb- 6C4.. choosing 4 black socks out of avail 6= 15.. 2)wwww- same as 1)-15 3)bbbw-6C3*6C1=120 4)wwwb-same as 3)=120 5)wwbb-6C2*6C2=15*15=225

now total ways =15+15+120+120+225=495.. the wwbb way=225 so prob=225/495=5/11.. hope the concept was clear..[/quote]

gmatclubot

Re: A drawer has six loose blue socks and six loose white socks.
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