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# A drawer contains 6 socks. If two socks are randomly

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Joined: 12 Sep 2015
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A drawer contains 6 socks. If two socks are randomly  [#permalink]

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14 Nov 2016, 08:12
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Difficulty:

95% (hard)

Question Stats:

28% (02:15) correct 72% (01:57) wrong based on 207 sessions

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A drawer contains 6 socks. If two socks are randomly selected without replacement, what is the probability that both socks will be black?

(1) The probability is less than 0.3 that the first sock selected will be black.
(2) The probability is greater than 0.4 that both socks will be white.

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A drawer contains 6 socks. If two socks are randomly  [#permalink]

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16 Nov 2016, 03:40
6
S1 - Probability of selecting 1 black sock from 6 socks =$$\frac{bC1}{6C1}$$=$$\frac{b}{6}$$<0.3
where b is no. of black socks.
i.e. b<1.8 i.e. b can be 0 or 1.
In either case the probability is 0. So sufficient.

S2 - Probability of both socks white >0.4
i.e.$$\frac{wC2}{6C2}$$ >0.4 where w is no. of white socks
$$\frac{w(w-1)}{30}$$> 0.4
therefore, w>4 i.e. no. of white socks is 5 or 6.
In either case, probability of picking 2 black socks is 0. So, sufficient.

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Re: A drawer contains 6 socks. If two socks are randomly  [#permalink]

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14 Nov 2016, 15:10
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GMATPrepNow wrote:
A drawer contains 6 socks. If two socks are randomly selected without replacement, what is the probability that both socks will be black?

(1) The probability is less than 0.3 that the first sock selected will be black.
(2) The probability is greater than 0.4 that both socks will be white.

* Kudos for all correct solutions

The number of black socks and white socks have to be integers
let's call The number of black socks b and The number of white socks w

(1) The probability is less than 0.3 that the first sock selected will be black.
it means that $$\frac{b}{6}<0.3$$ ===> $$b<1.8$$=====> $$b<=1$$
Thus, there is not possibility to selected without replacement 2 black socks because there is only one of those or none of those.

probability=0 SUFFICIENT

(2) The probability is greater than 0.4 that both socks will be white.
$$\frac{w}{6}*(w-1)/5>0.4$$ w*(w-1)>12 the lowest integer for w is 5

Thus there are 1 or less black socks
again there is not possibility to selected without replacement 2 black socks

probability=0 SUFFICIENT

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Re: A drawer contains 6 socks. If two socks are randomly  [#permalink]

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20 Jul 2017, 05:18
[quote="GMATPrepNow"]A drawer contains 6 socks. If two socks are randomly selected without replacement, what is the probability that both socks will be black?

(1) The probability is less than 0.3 that the first sock selected will be black.
(2) The probability is greater than 0.4 that both socks will be white.

* Kudos for all correct solutions

1) P(B)< 0.3
total socks=6
so, 0.3 x 6 = 1.8 ( black is less than this, since p(b) < 0.3)
black =1
1/6 x 0/5 = 0 sufficient

2) p(w and w)> 0.4
0.4x6 = 2.4 ( in total it must be greater than this, so it will be either 3,4,5 or even 6) but there will be only one value that satisfies the equation.
check values :
lets say : p(w and w) =3
3/6 x 2/5 = 1/5 (0.2,not good)
try p(w and w) =4
4/6 x 3/5 = 2/5 (= 0.4) where p (w and w ) must be greater than 0.4. not good
so either w=5 or w=6
both ways p(b and b) =0
sufficient

Ans : D
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Re: A drawer contains 6 socks. If two socks are randomly  [#permalink]

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25 Jul 2017, 03:20
Bunuel - Shouldn't the answer should be A. We do not have any information as to how many black socks are among the 6 (if any exist at all).
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Re: A drawer contains 6 socks. If two socks are randomly  [#permalink]

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30 Jul 2017, 17:48
2
GMATPrepNow wrote:
A drawer contains 6 socks. If two socks are randomly selected without replacement, what is the probability that both socks will be black?

(1) The probability is less than 0.3 that the first sock selected will be black.
(2) The probability is greater than 0.4 that both socks will be white.

We are given that a drawer contains 6 socks. If we let b = the number of black socks, we see that the probability of a black sock on the first pick is b/6. Since there is no replacement, the probability of a black sock on the second pick is (b - 1)/5. We need to determine the product of (b/6) x (b-1)/5.

Statement One Alone:

The probability is less than 0.3 that the first sock selected will be black.

Using the information in statement one, we have:

b/6 < 3/10

10b < 18

b < 1.8

We see that there is at most 1 black sock. Thus, the probability of pulling 2 black socks in two picks is equal to 0. Statement one alone is sufficient to answer the question.

Statement Two Alone:

The probability is greater than 0.4 that both socks will be white.

Suppose that there were exactly 4 white socks in the drawer. In this case, the probability of drawing 2 white socks would be 4/6 x 3/5 = 12/30 = 2/5 = 0.4. Since the probability of getting 2 white socks is greater than 0.4, there must be more than 4 white socks in the drawer; therefore, there can be at most 1 black sock in the drawer. The probability of pulling 2 black socks is 0. Statement two alone is sufficient to answer the question.

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Re: A drawer contains 6 socks. If two socks are randomly  [#permalink]

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19 Sep 2018, 21:41
Awesome, awesome question.Will make many a mathematically inclined people just kick themselves after seeing the answer.Not so tough but thoroughly enjoyed this problem....
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Re: A drawer contains 6 socks. If two socks are randomly  [#permalink]

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23 Nov 2019, 21:28
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Re: A drawer contains 6 socks. If two socks are randomly   [#permalink] 23 Nov 2019, 21:28
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