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A drawer contains 36 socks, and 2 socks are selected at

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A drawer contains 36 socks, and 2 socks are selected at  [#permalink]

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New post 25 Oct 2009, 04:13
3
12
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A
B
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D
E

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Question Stats:

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A drawer contains 36 socks, and 2 socks are selected at random without replacement. What is the probability that both socks are black?

(1) The probability is 4/9 that the first sock is black.
(2) The number of white socks in the drawer is 4 more than the number of black socks.
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Re: prob1  [#permalink]

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New post 25 Oct 2009, 16:08
8
Economist wrote:
A drawer contains 36 socks, and 2 socks are selected at random without replacement. What is the probability that both socks are black?


(1) The probability is 4/9 that the first sock is black.

(2) The number of white socks in the drawer is 4 more than the number of black socks.


Total # of socks 36.

(1) P(B)=4/9 --> B=16 --> P(BB)=16/36*15/35=4/21. Sufficient.

(2) W=B+4, as we don't know whether drawer contains only white and black socks, or there are also some other color socks, we can not determine the # of B. not sufficient.

Answer: A.
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Re: prob1  [#permalink]

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New post 26 Oct 2009, 04:15
Hi,
Think that the ans is E
It is not mentioned that the socks are only B&W
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Re: prob1  [#permalink]

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New post 26 Oct 2009, 07:26
2
BG wrote:
Hi,
Think that the ans is E
It is not mentioned that the socks are only B&W
Regards


You are right, it's not mentioned and that is why (2) is not sufficient. But for (1): saying that the probability of drawing the FIRST black sock is 4/9 means that we have 4 chance out of 9 to have black OR as there are total of 36 socks, 16 chances out of 36. That's clearly gives us the NUMBER OF BLACK SOCKS in the drawer, even though we still don't know the number and colors of the other socks (well, there are total 36-16=20 others but we don't know their color). So (1) is still sufficient to determine the probability that both socks are black 4/9*15/35=4/21.

Answer: A.
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Re: A drawer contains 36 socks, and 2 socks are selected at  [#permalink]

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New post 21 Mar 2012, 11:43
I think it's D:
(1) is obvious
(2) The number of white socks in the drawer is 4 more than the number of black socks.
Let’s B be the number of black socks and W the number of white.
W = B +4 and we know 36 = B +W then W = 36 – B = B + 4
=> 2B = 32 => B =16 (Sufficient !)
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Re: A drawer contains 36 socks, and 2 socks are selected at  [#permalink]

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New post 21 Mar 2012, 11:56
FoudMine wrote:
I think it's D:
(1) is obvious
(2) The number of white socks in the drawer is 4 more than the number of black socks.
Let’s B be the number of black socks and W the number of white.
W = B +4 and we know 36 = B +W then W = 36 – B = B + 4
=> 2B = 32 => B =16 (Sufficient !)



But you are wrongfully assuming from the question that there are only B and W socks, which is not necessarily the case.

IMO : A
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Re: prob1  [#permalink]

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New post 01 Jul 2012, 07:44
Bunuel wrote:
Economist wrote:
A drawer contains 36 socks, and 2 socks are selected at random without replacement. What is the probability that both socks are black?


(1) The probability is 4/9 that the first sock is black.

(2) The number of white socks in the drawer is 4 more than the number of black socks.


Total # of socks 36.

(1) P(B)=4/9 --> B=16 --> P(BB)=16/36*15/35=4/21. Sufficient.
(2) W=B+4, as we don't know whether drawer contains only white and black socks, or there are also some other color socks, we can not determine the # of B. not sufficient.

Answer: A.


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Re: A drawer contains 36 socks, and 2 socks are selected at  [#permalink]

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New post 07 Jun 2013, 05:12
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Re: A drawer contains 36 socks, and 2 socks are selected at  [#permalink]

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New post 09 Jun 2013, 01:41
2
Let the number of black socks = b

So the question can be rephrased as b/36 * (b-1)/35 = ?


Stmt 1 : The probability of the 1st sock being black = b/36 = 4/9

Cross multiply and you get b = 16

Hence we know that the 16 is the total number of black socks

This information helps us calculate the exact probability of getting 2 black socks. Hence stmt 1 is sufficient.



Stmt 2 : Let the number of white socks be w hence, w = b+4

This information does not help us arrive at the exact probability of getting 2 black socks because we are not told whether the 36 socks consist of only black and white socks or whether there are more colours as well.

Since we can not derive exact information,this statement is insufficient.


Answer : A
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Re: A drawer contains 36 socks, and 2 socks are selected at  [#permalink]

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New post 02 Nov 2018, 01:13
A drawer contains 36 socks, and 2 socks are selected at random without replacement. What is the probability that both socks are black?

(1) The probability is 4/9 that the first sock is black.
(2) The number of white socks in the drawer is 4 more than the number of black socks.

From either statement, one can conclude that there are 16 black and 20 white socks. So the desired probability is 16/36 x 15/35

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Re: A drawer contains 36 socks, and 2 socks are selected at &nbs [#permalink] 02 Nov 2018, 01:13
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A drawer contains 36 socks, and 2 socks are selected at

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