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

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akhil911

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18 May 2014, 00:27

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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

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gmatacequants

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18 May 2014, 09:09

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akhil911

Hi Brunei,

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

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

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18 May 2014, 00:46

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akhil911

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

Answer: A.

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akhil911

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18 May 2014, 04:55

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Hi Brunei,

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

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Updated on: 18 May 2014, 21:53

Originally posted by shaderon on 18 May 2014, 21:13.
Last edited by shaderon on 18 May 2014, 21:53, edited 1 time in total.

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

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.

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18 May 2014, 21:26

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shaderon

akhil911

Can someone help me with the approach?

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

I'm not really sure how to progress after this.
I have to take into account the different ways this could happen, right?
How do I do that?

Is the approach OK?

Thanks in advance.

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

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18 May 2014, 22:16

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shaderon

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.

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03 Nov 2014, 09:40

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akhil911

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.

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10 Jun 2015, 06:39

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

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

Probability = 3/15 = 1/5

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22 Feb 2016, 23:51

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experts please tell whats wrong in this

6/6*1/5*4/4*1/3 *6|/2*2 =2/5

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09 Mar 2016, 13:43

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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!

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09 Mar 2016, 13:47

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AlexIV

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!

C stands for combinations:

Combinatorics Made Easy!

Theory on Combinations

DS questions on Combinations
PS questions on Combinations

Hope it helps.

AlexIV

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09 Mar 2016, 16:34

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Thank you Bunuel!!!

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28 Mar 2017, 02:10

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Can someone please explain in this format? :

[2/6*1/5*2/4*1/3]*3 = 1/30 is my answer...

How can i get a 6 in the numerator so that the answer becomes 1/5? What would the logic be?

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24 Apr 2017, 16:32

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Bunuel

akhil911

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

Answer: A.

Isn't it supposed to be \(C^4_6\) = picking 4 socks out of 6 (i.e 2 pairs)?

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07 Jun 2017, 22:54

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Hi Bunuel,

Why is the below approach incorrect?

6c1*1*4c1*1/(4!*6c4) Firstly, we are selecting one sock out of 6, then we can select second in just one way as we are assuming that it is the second sock of the pair and so on; since the combination can be any so divide by 4! and total cases are 6c4.

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08 Jun 2017, 12:01

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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.

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10 Jun 2017, 08:12

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akhil911

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

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Quote:

akhil911 wrote:
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

Plz let me know if my approach is wrong-

Total number of socks = 6

Total number of ways Sarah can pick up 4 socks = 6C4 = 15 ways

No. of ways he can pick matching shoes = no.of ways 2 red socks + 2 blue socks + 2 green socks

= 2C2 + 2C2 + 2C2

= 1 + 1 + 1

= 3 ways

Hence

P(Matching socks) =3/15 = 1/15