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

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 \(C^3_2\)

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:

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

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.






Posted from my mobile device

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.

Why isn't the numerator multiplied by 2, since there are 2! ways to arrange both pairs??

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.

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

chetan2u nick1816

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

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

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

Kritisood
EMPOWERgmatRichC explained it perfectly. You need to consider every possible group of to pairs. So you have to multiply by 3C2.

Another point i would like to add is that you don't need to multiply by 4!/2!*2!...You've already did the arrangements in the first part. That's the difference between your solution and Bunuel's. Bunuel consider only the selection thing.

Point is arrangements got cancelled out while finding the probability. Hence, if you consider only selection or both selection and arrangement, you will get the same answer.

To select \(4\) socks out of \(6\), which can be done by \(6C4\) = \(15\) ways.

Now, selection can be done either by RR-BB or RR-GG or BB-GG,

For RR-BB it is \(2C2\) x \(2C2\) x \(2C0\) = \(1\) [ \(2C0\) for no green];

Probability = \(\frac{1}{15}\);

Similarly for RR-GG and BB-GG we can get the probabilitiy of \(\frac{1}{15}\) from each.

Total probability = \(\frac{1}{15}\) + \(\frac{1}{15}\) + \(\frac{1}{15}\) = \(\frac{3}{15}\) = \(\frac{1}{5}\)

Given: In packing for a trip, Sarah puts three pairs of socks - one red, one blue, and one green - into one compartment of her suitcase.
Asked: 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?

Favorable ways = 3C2 = 3
Total ways = 6C4 = 15

Probability = Favorable ways/total ways = 3/15 = 1/5

IMO A

Picking up two pairs means leaving behind one exact pair.

Probability of picking two pairs = Probability of not picking one exact pair = 6/6 * 1/5 = 1/5

Ans: A

Hi Bunuel


WHat is wrong with my approach?

(6C1 X 1C1 X 4C1 X 1C1 ) *3C2/ 6C4
Selecting any sock from 6 socks, then making pair (1c1), then selecting 1 sock from 4 left socks and making one pair (1c1) ....then there are 3c2 ways to select the above condition (like RedRed BlueBlue, RedRed green green, BlueBlue, green green)

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

Let f represent the first color pulled and s be the second.

You can get this order:

F1F2S1S2

P(F) = 1 <--- First sock of any colour has a probability of 1
P(F2) = 1/5 <--- Second sock
P(S1) = 1/2
P(S2) = 1/3

1 x 1/5 x 1/2 x 1/3 = 2/60

Next, account for order

4! / 2! x 2! = 6

6 x 2/60 = 12/60 = 1/5

A.
_________________

anubhav802
by (6c1x 1c1 x 4c1 x 1c1) you have fixed the position of a particular sock you will have to divide it by 4! to get the required answer.

Probability of 2 matching pairs = Probability of getting 2 of one color and 2 more of another color
Number of ways of choosing the colors = 3C2 = 3 ways = Number of favorable cases
Total number of cases = 6C4 = 15 (choosing any 4 out of the 6)

Probability = 3/15 = 1/5
Answer A
_________________