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

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

Hope it helps.

Hi VedantChourey, I think you're on the right track if you wanted to use the probability method on this question.
However, you've only considered one case, that is Blue-White-Blue-White. You must consider all cases if you want to use this method. Keep in mind that it can be time consuming (it took me over 2min to come up with the cases and write out the formulas).

Double your first equation as the option of WWBB exists, giving us \(\frac{10}{66}\).
BWWB results in \(\frac{6}{12}*\frac{6}{11}*\frac{5}{10}*\frac{5}{9}\), giving another \(\frac{5}{66}\) or \(\frac{10}{66}\) if we consider the reverse (WBBW).
Finally, we have BWBW, which is the same formula as above, doubled to \(\frac{10}{66}\) for the reverse (WBWB).

In total, that is \(\frac{10}{66}\) three times, which gives us \(\frac{10}{66}*3=\frac{5}{11}\). Option D.

Hope that helps!

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

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

Hope it helps.

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)

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

[arent we drawing 1 card at a time and not 4 cards at a tiime ]

Regards,

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

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.)

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.

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?

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

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

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..

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..[/quote]

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?

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?