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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. 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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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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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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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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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]

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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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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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New post 26 May 2013, 23:48
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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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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New post 27 May 2013, 02:59
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.
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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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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..
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Re: A drawer has six loose blue socks and six loose white socks. [#permalink]

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New post 13 Feb 2015, 05:16
2(6c4 + 6c3*6c1)/12c4......5/11 answer
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New post 12 Dec 2015, 06:11
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.

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

Answer: D.
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New post 12 Dec 2015, 06:35
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.

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

Answer: D.


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..
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New post 12 Dec 2015, 06:44
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.

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

Answer: D.


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

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New post 12 Nov 2016, 11:56
Bunuel wrote:
kv18 wrote:

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

Hope it helps.


I think this question coming I believe from Kaplan's book is not clearly written, the answer 5/11 is only correct, if it really matters that I pull out a pair. As it was noted earlier, the case is that 4 socks are drawn (no matter simultaneously or not) and there are only 5 possible scenarios (W- White, B - Black, order does not matter): WWWW, BBBB, WWWB, BBBW, BBWW. Why in the world, the probability is not simply 1/5?
The other point is that 5/11 = 45% in other words, it is almost 50/50 chance of getting socks right, which looks strange from common sense view..Suppose you have 4 hands and one can catch only one sock you put your hands into a drawer with 12 socks laying in any order, do you really have almost 50/50 chance to pull out 2 pairs?
A drawer has six loose blue socks and six loose white socks.   [#permalink] 12 Nov 2016, 11:56
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