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Re: M0206 [#permalink]
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04 Nov 2014, 02:06
the way i understood the question is that we have "pair" of socks" , while we are picking individual socks .... questions should not be this vague if you ask me .



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Re: M0206 [#permalink]
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Re: M0206 [#permalink]
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04 Nov 2014, 10:43
Bunuel wrote: elitherare wrote: the way i understood the question is that we have "pair" of socks" , while we are picking individual socks .... questions should not be this vague if you ask me . Thank you for your feedback. Edited as: There are 5 pairs of white, 3 pairs of black and 2 pairs of grey socks in a drawer. If four individual socks are picked at random what is the probability of getting at least two socks of the same color? Is this better? Dear Sir , the way it is phrased right now the answers ought to be change because we have 10 white socks and 6 black socks (5 and 3 pairs respectively) and we are picking 4 out of them In my opinion the question should be rephrased as "there are 5 white socks and 3 black socks , if 4 socks are picked etc..." in case u insist in using the word "pairs" , then it must be used before each number "5 pairs of white" , 3 "pairs" of black , if 4 "pairs" are chosen randomly.



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Re: M0206 [#permalink]
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05 Nov 2014, 04:48
elitherare wrote: Bunuel wrote: elitherare wrote: the way i understood the question is that we have "pair" of socks" , while we are picking individual socks .... questions should not be this vague if you ask me . Thank you for your feedback. Edited as: There are 5 pairs of white, 3 pairs of black and 2 pairs of grey socks in a drawer. If four individual socks are picked at random what is the probability of getting at least two socks of the same color? Is this better? Dear Sir , the way it is phrased right now the answers ought to be change because we have 10 white socks and 6 black socks (5 and 3 pairs respectively) and we are picking 4 out of them In my opinion the question should be rephrased as "there are 5 white socks and 3 black socks , if 4 socks are picked etc..." in case u insist in using the word "pairs" , then it must be used before each number "5 pairs of white" , 3 "pairs" of black , if 4 "pairs" are chosen randomly. No. In that sense the question is fine. There are 5 pairs of white (so 10), 3 pairs of black (so 6) and 2 pairs of grey (so 4) socks in a drawer. If four individual socks are picked at random what is the probability of getting at least two socks of the same color? Ans the answer is 100%.
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Re: M0206 [#permalink]
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20 Mar 2015, 07:11
Dear Sir, If there are 10 white, 6 black and 4 green socks and if 4 socks are picked up individually, then for instance all the 4 may be white, which makes the probability of at least 2 same colored socks as 0.



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20 Mar 2015, 08:07



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Re: M0206 [#permalink]
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20 Mar 2015, 08:55
Bunuel,
Thanks for the question. If it were phrased differently and we only picked 3 individual socks instead of 4 individual socks, how would we approach the question?



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Re: M0206 [#permalink]
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20 Mar 2015, 09:19
I guess that must be calculated as below, p(at least 2 socks same colour)=1P(all 3 are different colors) 110x6x4/ 20C3 124/114 =45/57
However I am not too sure. Bunuel, kindly confirm. Thanks for your previous reply.



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Re: M0206 [#permalink]
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20 Mar 2015, 09:44
MEOWSER wrote: Bunuel,
Thanks for the question. If it were phrased differently and we only picked 3 individual socks instead of 4 individual socks, how would we approach the question? the recommended way would be.. in this case you will have to find the probability of all three being different... since there are three different socks , the ways to pick up three amongst th three would be 3!=6... now the probability in each way=\(\frac{10}{20}*\frac{6}{19}*\frac{4}{18}\)=\(\frac{2}{57}\)... so for 6 ways=6*\(\frac{2}{57}\)= \(\frac{12}{57}\).. so the prob of not having all different or in other words having atleast two of same colour= 1\(\frac{12}{57}\)=\(\frac{45}{57}\)
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Re: M0206 [#permalink]
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22 Mar 2015, 03:44
chetan2u wrote: MEOWSER wrote: Bunuel,
Thanks for the question. If it were phrased differently and we only picked 3 individual socks instead of 4 individual socks, how would we approach the question? the recommended way would be.. in this case you will have to find the probability of all three being different... since there are three different socks , the ways to pick up three amongst th three would be 3!=6... now the probability in each way=\(\frac{10}{20}*\frac{6}{19}*\frac{4}{18}\)=\(\frac{2}{57}\)... so for 6 ways=6*\(\frac{2}{57}\)= \(\frac{12}{57}\).. so the prob of not having all different or in other words having atleast two of same colour= 1\(\frac{12}{57}\)=\(\frac{45}{57}\) How are there 6 ways to pick up the socks? The selection order shouldn't matter, so it's a combination: 3C3 = 1, and 1  2/57 = prob of having at least two of the same. Please advise MEOWSER or Bunuel.



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Re: M0206 [#permalink]
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12 Oct 2015, 15:29
Bunuel wrote: There are 5 pairs of white, 3 pairs of black and 2 pairs of grey socks in a drawer. If four individual socks are picked at random what is the probability of getting at least two socks of the same color?
A. \(\frac{1}{5}\) B. \(\frac{2}{5}\) C. \(1\) D. \(\frac{3}{5}\) E. \(\frac{4}{5}\) Bunuelnaturally in all the probability and combinations problems, a condition with or without replacement is mentioned which plays a key role in deciding the answer. Please consider this aspect.
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Re M0206 [#permalink]
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12 May 2016, 03:25
I think this is a highquality question and I agree with explanation.



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Re M0206 [#permalink]
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16 Jul 2016, 10:21
I think this is a highquality question and I agree with explanation.



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samners wrote: chetan2u wrote: MEOWSER wrote: Bunuel,
Thanks for the question. If it were phrased differently and we only picked 3 individual socks instead of 4 individual socks, how would we approach the question? the recommended way would be.. in this case you will have to find the probability of all three being different... since there are three different socks , the ways to pick up three amongst th three would be 3!=6... now the probability in each way=\(\frac{10}{20}*\frac{6}{19}*\frac{4}{18}\)=\(\frac{2}{57}\)... so for 6 ways=6*\(\frac{2}{57}\)= \(\frac{12}{57}\).. so the prob of not having all different or in other words having atleast two of same colour= 1\(\frac{12}{57}\)=\(\frac{45}{57}\) How are there 6 ways to pick up the socks? The selection order shouldn't matter, so it's a combination: 3C3 = 1, and 1  2/57 = prob of having at least two of the same. Please advise MEOWSER or Bunuel. Can someone please clarify whether we have to multiply with 3!=6 In my opinion we have to multiple with 3! precisely because the order does not matter. If we don't multiple with 3!, we will only have the probability for one specific event (1. white sock, 2. black sock, 3. green sock). Kind regards.



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Re M0206 [#permalink]
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27 Sep 2016, 23:21
I think this is a highquality question and I agree with explanation.



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Re M0206 [#permalink]
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06 Oct 2016, 07:04
I think this is a poorquality question and I don't agree with the explanation. wrong question



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Re: M0206 [#permalink]
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06 Oct 2016, 12:32
Can anyone share any other method apart from the official solution by Bunuel to this question.



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Re M0206 [#permalink]
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12 Nov 2017, 09:58
I think this is a poorquality question and I agree with explanation. This is a duplicate of M2610.










