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

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Re: In packing for a trip, Sarah puts three pairs of socks - one  [#permalink]

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New post 03 Jul 2017, 01:32
Bunuel wrote:
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

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


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

Answer: A.


Bunuel
Aren't any socks of the same color identical?
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In packing for a trip, Sarah puts three pairs of socks - one  [#permalink]

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New post 29 Jul 2017, 09:41
Bunuel wrote:
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

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


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

Answer: A.


Why are you choose formulas upside down. I have never seen it presented in the other way. Also, it seems a little odds that we can use two different formulas to calculate the answer. 2 pairs out of 3 divided by individual socks.

I like the OA explanation better that of the 2 remaining socks not taken we have 1 sock that could be any color. And given it is red, blue or green what is the odds the other one is red blue or green.

What is the odds that this matches. So we know there are 6 stocks so 1/5 socks remain to match it. Or 1/5 match. And if the pair not chosen matches that means the other 2 also match. And as mentioned, I have always seen \(C^3_2\)
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Re: In packing for a trip, Sarah puts three pairs of socks - one  [#permalink]

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New post 07 Dec 2017, 14:44
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).

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Re: In packing for a trip, Sarah puts three pairs of socks - one  [#permalink]

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New post 23 Feb 2018, 22:47
You are pretty close .

6/6*1/5*4/4*1/3 now this can happen for any two pair of socks out of 3. Therefore multiply this by 3c2.




vipulgoel wrote:
experts please tell whats wrong in this

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


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Re: In packing for a trip, Sarah puts three pairs of socks - one  [#permalink]

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New post 12 May 2018, 07:42
VERITAS SOLUTION
A. 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.
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Re: In packing for a trip, Sarah puts three pairs of socks - one  [#permalink]

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New post 29 May 2018, 12:07
iamdp wrote:
Favorable outcomes > selecting 2 pairs out of 3 = 3c2 = 3

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

Probability = 3/15 = 1/5


Why isn't the numerator multiplied by 2, since there are 2! ways to arrange both pairs??
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Re: In packing for a trip, Sarah puts three pairs of socks - one  [#permalink]

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New post 21 Sep 2018, 04:40
5
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.
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Re: In packing for a trip, Sarah puts three pairs of socks - one &nbs [#permalink] 21 Sep 2018, 04:40

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