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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]  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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Posts: 30437
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Kudos [?]: 57610 [1] , given: 8819

Re: A drawer has six loose blue socks and six loose white socks. [#permalink]  23 Sep 2012, 07:59
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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.

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

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

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

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]  23 Sep 2012, 23:51
Can someone explain the maths in more detail - ie what C2/6 means?
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Posts: 30437
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Kudos [?]: 57610 [0], given: 8819

Re: A drawer has six loose blue socks and six loose white socks. [#permalink]  24 Sep 2012, 01:51
Expert's post
jordanshl wrote:
Can someone explain the maths in more detail - ie what C2/6 means?

Check here:
math-combinatorics-87345.html (Combinations)
math-probability-87244.html (Probability)

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

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

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,
Math Expert
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Posts: 30437
Followers: 5102

Kudos [?]: 57610 [0], given: 8819

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

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

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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Re: A drawer has six loose blue socks and six loose white socks. [#permalink]  27 May 2013, 02:08
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.)
Math Expert
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Posts: 30437
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Kudos [?]: 57610 [0], given: 8819

Re: A drawer has six loose blue socks and six loose white socks. [#permalink]  27 May 2013, 02:59
Expert's post
apd2006 wrote:
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.)

$$C^2_6$$ is picking 2 out of 6 without replacement.

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

Hope it helps.
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Re: A drawer has six loose blue socks and six loose white socks. [#permalink]  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?
Math Expert
Joined: 02 Sep 2009
Posts: 30437
Followers: 5102

Kudos [?]: 57610 [1] , given: 8819

Re: A drawer has six loose blue socks and six loose white socks. [#permalink]  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.
_________________
Intern
Joined: 17 Mar 2013
Posts: 7
Schools: ISB '15 (S), HKU '15
GMAT 1: 720 Q49 V39
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Kudos [?]: 0 [0], given: 39

Re: A drawer has six loose blue socks and six loose white socks. [#permalink]  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]  13 Feb 2015, 05:16
Re: A drawer has six loose blue socks and six loose white socks.   [#permalink] 13 Feb 2015, 05:16
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# A drawer has six loose blue socks and six loose white socks.

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