In packing for a trip, Sarah puts three pairs of socks - one

Question

In packing for a trip, Sarah puts three pairs of socks - one red, one blue, and one green - into one compartment of her suitcase. If she then pulls four individual socks out of the suitcase, simultaneously and at random, what is the probability that she pulls out exactly two matching pairs?

A. 1/5
B. 1/4
C. 1/3
D. 2/3
E. 4/5

I have not understood how to solve this question properly , can someone help me out here.
Kudos me if you like the post !!!

A. 1/5
B. 1/4
C. 1/3
D. 2/3
E. 4/5

gmatacequants · Accepted Answer

Hi Akhil,

If you look,the successful events are when Sarah pulls 2 matching pair of socks out of 3 matching pairs of socks, which can be done in 3C2 ways as the order is not important
So,3C2 = 3 ways are:
RRBB
RRGG
BBGG

Now,the ways in which 4 socks can be pulled out of 6 socks (3 pairs) = 6C4 = 15

So, probability = 3/15 = 1/5

Bunuel · Answer

\frac{C^2_3}{C^4_6}=\frac{3}{15}=\frac{1}{5} :  C^2_3  = picking 2 pairs out of 3 and  C^2_6  = picking 4 socks out of 6.

Answer: A.

akhil911 · Answer

Hi Brunei,

Can you help in explaining the answer in detail.
I still cant understand how you have plugged in the numbers.

shaderon · Answer

1 - P(Exactly 1 Pair)    =1 - ((\frac{2_C_1*2_C_1*2_C_2}{6_C_4})*3)   =1 - (\frac{12}{15})=   \frac{3}{15}   =\frac{1}{5}

I know this is a longer approach, but is this right?

NOTE:  I understood that this could happen in 3 ways. So, I multiplied by 3.

But I'm never completely sure how to always consider the NUMBER OF ARRANGEMENTS in certain questions.
I get confused when there are  different elements  to be considered and when there are  similar elements .
Can someone help me with this?

Thanks in advance.

WoundedTiger · Answer

Hi,

I think  (\frac{2_C_1*2_C_1*2_C_2}{6_C_4})  represents a case where in 1 type of matching socks are pulled out but there 3 pairs and any one of them can be pulled out so I think it should be

1 - 3*(\frac{2_C_1*2_C_1*2_C_2}{6_C_4})

1-12/15 or 3/15 or 1/5

KarishmaB · Answer

Yes, but you need to be more specific about the kind of problems you are talking about. Small things change the entire question in P&C so let me know the exact questions that confuse you.

rajarshee · Answer

easier solution: inverse the problem. Sarah picking up 2 pair of matching socks and keeping one pair in the bag is equivalent to Sarah picking up 1 pair of socks and keeping 2 pairs in the bag. Now let's see what's the probability of that happening when done at random. Sarah picks up the 1st one, can be any color and so probability = 1. Sarah picks up the 2nd one, now for it to be of the same color as the 1st one, she has to pick 1 exact piece out of 5, probability of which is 1/5. So option A.

iamdp · Answer

Favorable outcomes > selecting 2 pairs out of 3 = 3c2 = 3

Total outcomes > selecting 4 socks randomly = 6c4 = 15

Probability = 3/15 = 1/5

AlexIV · Answer

Hi Guys, can somebody explain what does this C mean? where can I learn the theory? I have never seen this terminology in the books... It is killing me

Thanks!

Bunuel · Answer

C stands for combinations:

Combinatorics Made Easy!

Theory on Combinations

DS questions on Combinations 
 PS questions on Combinations

Hope it helps.

AlexIV · Answer

Thank you Bunuel!!!

avpooja · Answer

i did it using probability directly
(6/6 * 1/5 * 4/4 * 1/3 ) * 3C2 ways = (1/15) * 3 = 1/5

6/6 - we can choose any of the 6 socks in the first one
1/5 - we should pick the exact pair of the first sock picked so only 1 way out of the remaining 5 
4/4 - again we can pick any of the remaining 4 socks
1/3 - we have to pick the exact pair of the sock picked.

3C2 - because we could have either red + blue pair or red+ green pair or blue+green pair so 3 ways.

JeffTargetTestPrep · Answer

We have three scenarios of two matching pairs: 1) a red pair and a blue pair; 2) a red pair and a green pair; 3) a blue pair and a green pair. Let’s start with the probability of selecting a red pair and a blue pair. To select a red pair and a blue pair is to select two red socks and two blue socks. So let’s assume the first two socks are red and the last two socks are blue; the probability of selecting these socks in that order is:
 
P(R, R, B, B) = 2/6 x 1/5 x 2/4 x 1/3 = 1/6 x 1/5 x 1/3 = 1/90.
 
However, the two red socks and the two blue socks, in any order, can be selected in 4!/(2! x 2!) = 24/4 = 6 ways. Thus, the probability of two red socks and two blue socks is:
 
P(2R and 2B) = 1/90 x 6 = 6/90 = 1/15. 
 
Using similar logic, we see that the probability of pulling a red pair and a green pair is 1/15, and so is the probability of pulling a blue pair and a green pair. Thus, the total probability is:
 
1/15 + 1/15 + 1/15 = 3/15 = 1/5.
 
Alternate Solution:
 
From a total of 6 socks, two pairs, i.e., 4 socks, can be pulled in 6C4 = 6!/(4! 2!) = (6 x 5)/2 = 3 x 5 = 15 ways.
 
Three of these choices contain two matching pairs, namely: 1) a red pair and a blue pair, 2) a blue pair and a green pair; 3) a red pair and a green pair.
 
Therefore, the probability of pulling two matching pairs is 3/15 = 1/5.
 
Answer: A

EMPOWERgmatRichC · Answer

Hi All,

In probability questions, there are only 2 things that you can calculate: what you WANT and what you DON'T WANT.

(WANT) + (DON'T WANT) = 1

In this question, we can calculate the probability of what we DON'T WANT and subtract it from 1 to figure out the probability of what we do WANT.

The question asks for the probability of pulling 4 socks out that form 2 matching pairs. For this to occur, the two socks that are left would also form a matching pair. If the 2 leftover socks DO NOT form a matching pair, then the 4 socks that are pulled will NOT form 2 matching pairs.

Probability of 2 socks NOT forming a matching pair…

1st sock = 1 (any of the socks can be the first sock) 
2nd sock = 4/5 (since there's only one sock that matches the first sock).

Probability of NOT forming a pair with 2 socks: = 1 x 4/5 = 4/5 (which ALSO means a 4/5 chance of NOT having 2 matching pairs of 2 socks)

1 - 4/5 = 1/5 (meaning a 1/5 chance of having 2 matching pairs of 2 socks).

Final Answer:  A

GMAT assassins aren't born, they're made, 
Rich

HimanshuW11 · Answer

Veritas Prep Solution:

While it can be quite difficult to map out the possibilities for the four socks drawn, consider that drawing four socks will leave two remaining in the suitcase, so the problem can much more efficiently be solved by using the two "not drawn" socks. The only way that there will be two pairs drawn is if the two remaining socks are a pair. And the probability of one pair when picking two socks out of six would be 1 * (1/5), meaning that whatever the first sock is, the second will have to match it, and there's only one match left out of the remaining five socks. Therefore, the answer must be 1/5.

Kritisood · Answer

@experts could you please explain where I am going wrong?
6/6*1/5*4/4*1/3*4!/2!*2!

chetan2u   nick1816

EMPOWERgmatRichC · Answer

Hi Kritisood,

The first part of your calculation is fine:

(6/6)(1/5)(4/4)(1/3)

However, since there are 3 pairs of socks - and you need to account for every possible group of 2 pairs that could occur - you would have to multiple by 3c2...

(6/6)(1/5)(4/4)(1/3)(3!/2!1!) = 3/15 = 1/5

GMAT assassins aren't born, they're made,
Rich

Kritisood · Answer

thanks a lot Rick for the response. I was wondering why we wouldnt multiply it by 4!/2!*2! to "unarranged" since we have 2 socks of the same color and order doesn't matter eg RRBB

EMPOWERgmatRichC · Answer

Hi Kritisood,

The first part of your calculation actually deals with the individual socks, but the question requires that we consider the various PAIRS of socks that satisfy what we are looking for (and since there are 3 possible pairs that can be formed - and we're asked to form 2 of them, then you have to think of the math in those terms). In simple terms, any of the following combinations would fit what the question asks for:

Pair A and Pair B
Pair A and Pair C
Pair B and Pair C

That's 3 possible options, so we have to multiply that initial part of the calculation by 3 (which can be referred to as 3c2).

GMAT assassins aren't born, they're made,
Rich

In packing for a trip, Sarah puts three pairs of socks - one

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