A drawer contains 6 socks. If two socks are randomly

Question

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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souvonik2k · Accepted Answer

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.

Hence answer is D.

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damybox · Answer

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 answer D

hanyhamdani · Answer

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

laxpro2001 · Answer

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).

ScottTargetTestPrep · Answer

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.

Answer: D

redskull1 · Answer

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....

immarenbe · Answer

I have the same question. KarishmaB Bunuel chetan2u pls help me out. Thank you so much

chetan2u · Answer

Hi

White socks as per statement II are 5 or 6, otherwise the probability of picking two white socks will be less than 0.4

This means the black socks at the max will be 1, and probability of picking two black socks when there is only one or 0 of them will be 0.
So statement II should also be sufficient

KarishmaB · Answer

Total 6 socks.

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

Probability of selecting a black sock < 3/10 
So out of 6, the number of black socks is less than (3/10) * 6 i.e. 1.8. Hence, there is at most 1 black sock (since number of black socks must be an integer). Then probability of selecting 2 black socks without replacement is 0. 
Sufficient alone.

(2) The probability is greater than 0.4 that both socks will be white. 
If out of 6, 4 socks were white, probability of selecting both white socks without replacement would be (4/6)*(3/5) = 2/5 = 0.4
Since the probability of selecting both white is greater than 0.4, it means we have at least 5 white socks. If that is the case, then number of black socks is 1 or 0. 
So probability of selecting both black socks = 0
Sufficient alone

Answer (D)

immarenbe

NCC · Answer

Dear souvonik2k , My solution for statement 1 is: B = black balls P(B X ) < 0.3 BC1/ 6C1 X (6-B)C1/5 < 0.3 B/6 X (6-B)/5 < 0.3 B(6-B) < 9 -B^2 + 6B - 9 < 9 B(B-3)-3(B-3) < 9 solving the quadratic --> B > 3; so B = 4,5,6 why is it wrong?

bumpbot · Answer

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A drawer contains 6 socks. If two socks are randomly

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A drawer contains 6 socks. If two socks are randomly

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