A drawer has six loose blue socks and six loose white socks.

Question

A drawer has six loose blue socks and six loose white socks. If four socks are removed from the drawer at random and without replacement. What is the probability that one pair of each color was selected?

A. 2/33
B. 5/66
C. 5/33
D. 5/11
E. 1/2

A. 2/33
B. 5/66
C. 5/33
D. 5/11
E. 1/2

Bunuel · Accepted Answer

So, we want the probability of removing 2 blues socks out of 6 and 2 white socks out of 6, while removing 4 socks out of 12.

P=\frac{favorable}{total}=\frac{C^2_{6}*C^2_{6}}{C^4_{12}}=\frac{5}{11} .

Answer: D.

solo1234 · Answer

thanks u in advance. 
My book gives wrong anwers for this question.

Arthur1000 · Answer

Can someone explain the maths in more detail - ie what C2/6 means?

Bunuel · Answer

Check here: math-combinatorics-87345.html (Combinations) math-probability-87244.html (Probability) Hope it helps.

vomhorizon · Answer

No of Loose blue socks = 6 (3 pairs)
No. of Loose white Socks = 6 (3 pairs)
total no. of blue and white socks = 12 
Total no. of socks to be selected = 4

So we have C (12,4) = no of ways the socks can be selected = 495

No of ways One pair of blue socks is selected ( 2 blue socks) , C (6,2) , No of ways one pair of white socks can be selected C (6,2) ..Because we have to find a scenario where EXACTLY one pair of Blue socks and ONE pair of WHITE socks is selected we will multiply the two .. ie 15 x 15 ..

Filling the information in the Probability formula we get P (A) = (15 x 15) / 495 = 5 : 11 (D)

apd2006 · Answer

Hi, i was going through this question and tried a different approach and thus my answer differs ..please correct 
me if iam wrong..

4 cards are selected at random without replacement.
1st card then 2nd card then 3rd and then 4th card.
and we have 6B and 6W

So P(selecting 2 cards of same color one by one and the 2 cards of other same color)=6*5*6*5/12*11*10*9 = 5/66

Regards,

Bunuel · Answer

Mathematically the probability of picking 4 socks simultaneously, or picking them one at a time (without replacement) is the same.

apd2006 · Answer

But our answers differ. Which approach/answer is correct?
In your approach to solution ,you are replacing (taking number of different combinations of 4 at a time, that means you are replacing the socks back,
otherwise why is the count of socks not decreasing? and this is done in both -while calculating the favorable and total outcomes.)

Bunuel · Answer

Your approach is wrong. C^2_6 is picking 2 out of 6 without replacement. Check here for more: math-combinatorics-87345.html Hope it helps.

kv18 · Answer

Hi, I was trying through the probability approach:

the probability of selecting 2 blue and 2 white socks in the order (12/12 * 5/11 * 6/10 * 5/9) = 5/33

No. of possible orders = 4!/(2! * 2!) = 6 and all have same probability

p = 6 * 5 /33 = 10/11.

Why is this going wrong?

Bunuel · Answer

It should be (6/12 * 5/11 * 6/10 * 5/9) * 4!/(2! * 2!) = 5/11.

Hope it helps.

kv18 · Answer

Ohh...thank you so much Bunuel...looks like I need to get my basics right..

Cadaver · Answer

2(6c4 + 6c3*6c1)/12c4......5/11 answer

rsaahil90 · Answer

Hello  Bunuel

If the socks are identical, then the number of ways of selecting a sock in any turn would be 2 (either a W or B). If they form a pair they will be identical?

IMO - all possible selections are - {bbbb, wwww, bbbw, wwwb, wwbb} out of which wwbb is what we require ~ hence answer should be 1/5?

For e.g. if white and blue here would be boys and girls and then we are asked to find a team which has exactly 2 boys and 2 girls - In this case, the solution you mentioned should be valid.

Can you please explain.

Thank you

chetan2u · Answer

Hi  rsaahil90 ,
you may be correct in the 5 types of combination ..
however you have 6 pairs from which you have to choose these combinations and each combination does not have same weightage...
lets see this question only..
combinations ..
1) bbbb- 6C4.. choosing 4 black socks out of avail 6= 15..
2)wwww- same as 1)-15
3)bbbw-6C3*6C1=120
4)wwwb-same as 3)=120
5)wwbb-6C2*6C2=15*15=225

now total ways =15+15+120+120+225=495..
the wwbb way=225
so prob=225/495=5/11..
hope the concept was clear..

rsaahil90 · Answer

Hello Chetan

This is my whole concern - if all six of the socks are identical then there is just one way to choose 2 or 3 or 6 socks from that color. For e.g. if you have 5 red colored balls, in how many ways you can choose three i.e. 1 way as all the red balls are identical. If in this example balls would have been boys, then 5C3 ways are possible but not in case of red balls or socks in the concerned question

Thanks

Hi  rsaahil90 ,
you may be correct in the 5 types of combination ..
however you have 6 pairs from which you have to choose these combinations and each combination does not have same weightage...
lets see this question only..
combinations ..
1) bbbb- 6C4.. choosing 4 black socks out of avail 6= 15..
2)wwww- same as 1)-15
3)bbbw-6C3*6C1=120
4)wwwb-same as 3)=120
5)wwbb-6C2*6C2=15*15=225

now total ways =15+15+120+120+225=495..
the wwbb way=225
so prob=225/495=5/11..
hope the concept was clear..

Erjan_S · Answer

I think this question coming I believe from Kaplan's book is not clearly written, the answer 5/11 is only correct, if it really matters that I pull out a pair. As it was noted earlier, the case is that 4 socks are drawn (no matter simultaneously or not) and there are only 5 possible scenarios (W- White, B - Black, order does not matter): WWWW, BBBB, WWWB, BBBW, BBWW. Why in the world, the probability is not simply 1/5? 
The other point is that 5/11 = 45% in other words, it is almost 50/50 chance of getting socks right, which looks strange from common sense view..Suppose you have 4 hands and one can catch only one sock you put your hands into a drawer with 12 socks laying in any order, do you really have almost 50/50 chance to pull out 2 pairs?

CEdward · Answer

Bunuel  I am wondering why you didn't consider the order of the socks here?

e.g. 6C2 x 6C2 x 4!/2!x2! <--- Would this have been incorrect?

VedantChourey · Answer

Hi Mod's or anyone, whats wrong with this approach

I SOLVED THIS QUESTION IN THIS WAY,

6/12*5/11*6/10*5/9

First getting any 6 blue ball out of 12, then getting 5 blue ball(because its mentioned no replacement) out of 11, then getting any 6 white ball out of 10 and at last getting any 5 white ball out of 9. It gave me 5/66.

Whats wrong with this approach.

A drawer has six loose blue socks and six loose white socks.

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

A drawer has six loose blue socks and six loose white socks.

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards