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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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Difficulty:

65% (hard)

Question Stats:

56% (02:58) correct
44% (01:57) wrong based on 123 sessions

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]
23 Sep 2012, 07:59

1

This post received KUDOS

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]
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.

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) _________________

"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]
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.

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.

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

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

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.

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