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# M02-06

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Math Expert
Joined: 02 Sep 2009
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15 Sep 2014, 23:17
2
15
00:00

Difficulty:

65% (hard)

Question Stats:

64% (01:18) correct 36% (01:59) wrong based on 154 sessions

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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. $$\frac{4}{5}$$

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15 Sep 2014, 23:17
Official Solution:

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. $$\frac{4}{5}$$

No formula is need to answer this question. The trick here is that we have only 3 different color socks but we pick 4 socks, which ensures that in ANY case we'll have at least one pair of the same color (if 3 socks we pick are of the different color, then the 4th sock must match with either of previously picked one). P=1.

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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 .
Math Expert
Joined: 02 Sep 2009
Posts: 50730

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04 Nov 2014, 04:23
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?
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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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Joined: 02 Sep 2009
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05 Nov 2014, 04:48
1
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?

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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.
Math Expert
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20 Mar 2015, 08:07
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.

If there are 4 white socks aren't at least 2 of them same color, white?
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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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Joined: 19 Feb 2014
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20 Mar 2015, 09:19
I guess that must be calculated as below,
p(at least 2 socks same colour)=1-P(all 3 are different colors)
1-10x6x4/ 20C3
1-24/114
=45/57

However I am not too sure. Bunuel, kindly confirm. Thanks for your previous reply.
Math Expert
Joined: 02 Aug 2009
Posts: 7044

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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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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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Joined: 18 Sep 2014
Posts: 1127
Location: India

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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. $$\frac{4}{5}$$

Bunuel

naturally in all the probability and combinations problems, a condition with or without replacement is mentioned which plays a key role in deciding the answer.

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Joined: 29 Nov 2015
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12 May 2016, 03:25
I think this is a high-quality question and I agree with explanation.
Manager
Joined: 23 Apr 2014
Posts: 63
Location: United States
GMAT 1: 680 Q50 V31
GPA: 2.75

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16 Jul 2016, 10:21
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 27 Jun 2015
Posts: 19

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31 Aug 2016, 03:39
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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Joined: 12 Apr 2016
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27 Sep 2016, 23:21
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 10 Mar 2016
Posts: 9
Location: India
Schools: ISB '20
GMAT 1: 640 Q49 V28
GPA: 2.51

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06 Oct 2016, 07:04
I think this is a poor-quality question and I don't agree with the explanation. wrong question
Current Student
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Posts: 865
Concentration: Strategy
Schools: Fisher '19 (M)
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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.
Intern
Joined: 22 Jan 2017
Posts: 34

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12 Nov 2017, 09:58
I think this is a poor-quality question and I agree with explanation. This is a duplicate of M26-10.
Re M02-06 &nbs [#permalink] 12 Nov 2017, 09:58

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# M02-06

Moderators: chetan2u, Bunuel

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