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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? A. 2/33 B. 5/66 C. 5/33 D. 5/11 E. 1/2
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Last edited by Bunuel on 23 Sep 2012, 09:00, edited 1 time in total.
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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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Re: A drawer has six loose blue socks and six loose white socks. [#permalink]
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22 Jul 2013, 03:37



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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. \(P=\frac{favorable}{total}=\frac{C^2_{6}*C^2_{6}}{C^4_{12}}=\frac{5}{11}\). 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]
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24 Sep 2012, 00:51
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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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)
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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. \(P=\frac{favorable}{total}=\frac{C^2_{6}*C^2_{6}}{C^4_{12}}=\frac{5}{11}\). 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]
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27 May 2013, 00:48
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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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.)



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Re: A drawer has six loose blue socks and six loose white socks. [#permalink]
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Re: A drawer has six loose blue socks and six loose white socks. [#permalink]
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22 Jul 2013, 01:34
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]
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22 Jul 2013, 05:00
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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13 Feb 2015, 06:16
2(6c4 + 6c3*6c1)/12c4......5/11 answer



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A drawer has six loose blue socks and six loose white socks. [#permalink]
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12 Dec 2015, 07:11
Hello BunuelIf 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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Re: A drawer has six loose blue socks and six loose white socks. [#permalink]
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12 Dec 2015, 07:35
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)bbbw6C3*6C1=120 4)wwwbsame as 3)=120 5)wwbb6C2*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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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. \(P=\frac{favorable}{total}=\frac{C^2_{6}*C^2_{6}}{C^4_{12}}=\frac{5}{11}\). 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)bbbw6C3*6C1=120 4)wwwbsame as 3)=120 5)wwbb6C2*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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12 Nov 2016, 12:56
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?




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