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

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A drawer has six loose blue socks and six loose white socks. [#permalink] New post 23 Sep 2012, 07:06
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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?

A. 2/33
B. 5/66
C. 5/33
D. 5/11
E. 1/2
[Reveal] Spoiler: OA

Last edited by Bunuel on 23 Sep 2012, 08:00, edited 1 time in total.
Edited the question.
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Re: A drawer has six loose blue socks and six loose white socks. [#permalink] New post 23 Sep 2012, 07:59
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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.

P=\frac{favorable}{total}=\frac{C^2_{6}*C^2_{6}}{C^4_{12}}=\frac{5}{11}.

Answer: D.
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Re: A drawer has six loose blue socks and six loose white socks. [#permalink] New post 22 Jul 2013, 02:37
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kv18 wrote:
Hi, I was trying through the probability approach:

the probability of selecting 2 blue and 2 white socks in the order (12/12 * 5/11 * 6/10 * 5/9) = 5/33

No. of possible orders = 4!/(2! * 2!) = 6 and all have same probability

p = 6 * 5 /33 = 10/11.

Why is this going wrong?


It should be (6/12 * 5/11 * 6/10 * 5/9) * 4!/(2! * 2!) = 5/11.

Hope it helps.
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Re: A drawer has six loose blue socks and six loose white socks. [#permalink] New post 23 Sep 2012, 18:05
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.

P=\frac{favorable}{total}=\frac{C^2_{6}*C^2_{6}}{C^4_{12}}=\frac{5}{11}.

Answer: D.


thanks u in advance.
My book gives wrong anwers for this question.
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Re: A drawer has six loose blue socks and six loose white socks. [#permalink] New post 23 Sep 2012, 23:51
Can someone explain the maths in more detail - ie what C2/6 means?
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Re: A drawer has six loose blue socks and six loose white socks. [#permalink] New post 24 Sep 2012, 01:51
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Re: A drawer has six loose blue socks and six loose white socks. [#permalink] New post 24 Sep 2012, 23:27
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)
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Re: A drawer has six loose blue socks and six loose white socks. [#permalink] New post 26 May 2013, 08:43
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.

P=\frac{favorable}{total}=\frac{C^2_{6}*C^2_{6}}{C^4_{12}}=\frac{5}{11}.

Answer: D.



Hi, i was going through this question and tried a different approach and thus my answer differs ..please correct
me if iam wrong..

4 cards are selected at random without replacement.
1st card then 2nd card then 3rd and then 4th card.
and we have 6B and 6W

So P(selecting 2 cards of same color one by one and the 2 cards of other same color)=6*5*6*5/12*11*10*9 = 5/66

[arent we drawing 1 card at a time and not 4 cards at a tiime ]

Regards,
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Re: A drawer has six loose blue socks and six loose white socks. [#permalink] New post 26 May 2013, 23:48
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.

P=\frac{favorable}{total}=\frac{C^2_{6}*C^2_{6}}{C^4_{12}}=\frac{5}{11}.

Answer: D.



Hi, i was going through this question and tried a different approach and thus my answer differs ..please correct
me if iam wrong..

4 cards are selected at random without replacement.
1st card then 2nd card then 3rd and then 4th card.
and we have 6B and 6W

So P(selecting 2 cards of same color one by one and the 2 cards of other same color)=6*5*6*5/12*11*10*9 = 5/66

[arent we drawing 1 card at a time and not 4 cards at a tiime ]

Regards,


Mathematically the probability of picking 4 socks simultaneously, or picking them one at a time (without replacement) is the same.
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: A drawer has six loose blue socks and six loose white socks. [#permalink] New post 27 May 2013, 02:08
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.)
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Re: A drawer has six loose blue socks and six loose white socks. [#permalink] New post 27 May 2013, 02:59
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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.

Check here for more: math-combinatorics-87345.html

Hope it helps.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: A drawer has six loose blue socks and six loose white socks. [#permalink] New post 22 Jul 2013, 00:34
Hi, I was trying through the probability approach:

the probability of selecting 2 blue and 2 white socks in the order (12/12 * 5/11 * 6/10 * 5/9) = 5/33

No. of possible orders = 4!/(2! * 2!) = 6 and all have same probability

p = 6 * 5 /33 = 10/11.

Why is this going wrong?
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Re: A drawer has six loose blue socks and six loose white socks. [#permalink] New post 22 Jul 2013, 04:00
Bunuel wrote:
kv18 wrote:
Hi, I was trying through the probability approach:

the probability of selecting 2 blue and 2 white socks in the order (12/12 * 5/11 * 6/10 * 5/9) = 5/33

No. of possible orders = 4!/(2! * 2!) = 6 and all have same probability

p = 6 * 5 /33 = 10/11.

Why is this going wrong?


It should be (6/12 * 5/11 * 6/10 * 5/9) * 4!/(2! * 2!) = 5/11.

Hope it helps.


Ohh...thank you so much Bunuel...looks like I need to get my basics right..
Re: A drawer has six loose blue socks and six loose white socks.   [#permalink] 22 Jul 2013, 04:00
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