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Probability - 2 Pairs of Socks

This topic has 9 expert replies and 5 member replies
bml1105 Master | Next Rank: 500 Posts
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Probability - 2 Pairs of Socks

Post Sun May 25, 2014 4:02 pm
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

OA: A

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Zoser Junior | Next Rank: 30 Posts
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Post Sat May 27, 2017 10:11 am
Quote:
The first sock selected can be any color.
Of the 5 remaining socks, 3 will be selected.
Isn't the second selection to match the first color is 1/5 not 3/5?

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bml1105 Master | Next Rank: 500 Posts
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Post Thu May 29, 2014 4:42 pm
Thank you everyone! It finally clicked.

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Post Fri Dec 15, 2017 6:59 am
bml1105 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
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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YTarhouni Newbie | Next Rank: 10 Posts Default Avatar
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Post Sun Sep 03, 2017 2:11 pm
Math:
P(R&R)=P(B&B)=P(G&G)=2/6*1/5=2/30
Total probability=P(R&R+P(B&B)+P(G&G)=2/30+2/30+2/30=6/30=1/5

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Matt@VeritasPrep GMAT Instructor
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Post Mon Jun 05, 2017 10:47 pm
Zoser wrote:
Quote:
The first sock selected can be any color.
Of the 5 remaining socks, 3 will be selected.
Isn't the second selection to match the first color is 1/5 not 3/5?
That's if we want to get a match immediately, e.g. RR. If we just want a match at some point, we could have RWWR, RWRW, RRWW, etc. As long as one of the three remaining socks matches the first R, we're good.

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Post Sat May 27, 2017 12:15 pm
Zoser wrote:
Quote:
The first sock selected can be any color.
Of the 5 remaining socks, 3 will be selected.
Isn't the second selection to match the first color is 1/5 not 3/5?
For two matching pairs to be selected, the first sock selected must match one of the other 3 socks selected.
Once the first sock has been selected, 3 of the remaining 5 socks will then be selected.
Thus, the probability that this group of 3 will provide a match for the first sock is 3/5.

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Post Tue May 27, 2014 10:56 am
One more approach:

The first sock selected can be any color.
Of the 5 remaining socks, 3 will be selected.
Thus, the probability that a sock will be selected to match the first = 3/5.
The third sock selected can be any color.
Of the 3 remaining socks, 1 will be selected.
Thus, the probability that the last sock selected will match the third = 1/3.
To combine these probabilities, we multiply:
3/5 * 1/3 = 1/5.

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Post Tue May 27, 2014 9:07 am
bml1105 wrote:
I think where I get confused with this question is that if there are three pairs of socks (6 total socks) and 4 are randomly selected, a pair of socks must be chosen. It doesn't matter which the pair are (red, green or blue), but that leaves one pair out of the loop for being chosen and limits are choices for the last two socks to 4 socks.

I took that pair out and then tried to do the problem both ways: (1) (probability of one sock pulled and the second sock matching) x (1 pair that was definitely pulled) and (2) (1 - probability of one sock pulled and the second sock not matching) x (1 pair that was definitely pulled). Both ways I got (C) 1/3

Should it be a general rule that I never take out the pair that is definitely made?
In taking out a matching pair, you imply that the first 2 socks selected will DEFINITELY form a matching pair.
Not so.
In fact, it is far more likely that the first 2 socks selected will NOT form a matching pair.
Since you take it as a GIVEN that the first 2 socks will form a matching pair, you INCREASE the odds of selecting two matching pairs.
The result is that your answer (1/3) is greater than the ACTUAL probability of selecting 2 matching pairs (1/5).

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Post Tue May 27, 2014 8:40 am
bml1105 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

We can also use counting methods to solve this question.

IMPORTANT: If there are 2 pairs among the 4 SELECTED socks, then the 2 socks that are NOT SELECTED must also form a pair.
So, P(2 pairs among the 4 selected socks) = P(2 NOT SELECTED socks are paired)

P(2 NOT SELECTED socks are paired) = (# outcomes in which the 2 socks are paired)/(total # of ways to have 2 not selected socks)

# outcomes in which the 2 socks are paired
We have have 2 red socks, 2 blue socks or 2 green socks.
So, there are 3 outcomes in which the 2 socks are paired

total # of ways to have 2 not selected socks
There are 6 socks altogether, and we want to not select 2 of them.
Since the order doesn't matter here, we can use combinations.
We can choose 2 socks from 6 socks in 6C2 ways (15 ways)

Aside: If anyone is interested, we have a free video on calculating combinations (like 6C2) in your head: http://www.gmatprepnow.com/module/gmat-counting?id=789

So, P(2 NOT SELECTED socks are paired) = 3/15
= 1/5
= A

Cheers,
Brent

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Post Tue May 27, 2014 7:13 am
bml1105 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

IMPORTANT: As Rich has noted, if there are 2 pairs among the 4 SELECTED socks, then the 2 socks that are NOT SELECTED must also form a pair.
So, P(2 pairs among the 4 selected socks) = P(2 NOT SELECTED socks are paired)

P(2 NOT SELECTED socks are paired) = P(1st not selected sock is any color AND 2nd not selected sock matches the 1st)
= P(1st not selected sock is any color) x P(2nd not selected sock matches the 1st)
= 1 x 1/5
= 1/5
= A

Cheers,
Brent

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bml1105 Master | Next Rank: 500 Posts
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Post Mon May 26, 2014 7:44 pm
I think where I get confused with this question is that if there are three pairs of socks (6 total socks) and 4 are randomly selected, a pair of socks must be chosen. It doesn't matter which the pair are (red, green or blue), but that leaves one pair out of the loop for being chosen and limits are choices for the last two socks to 4 socks.

I took that pair out and then tried to do the problem both ways: (1) (probability of one sock pulled and the second sock matching) x (1 pair that was definitely pulled) and (2) (1 - probability of one sock pulled and the second sock not matching) x (1 pair that was definitely pulled). Both ways I got (C) 1/3

Should it be a general rule that I never take out the pair that is definitely made?

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Post Mon May 26, 2014 11:07 am
Hi bml1105,

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

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Post Mon May 26, 2014 1:52 am
RR, GG, BB ==> 4 SOCKS

Probability of Two pairs = RRGG + RRBB + GGBB

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

18 * 1/3 * 1/5 * 1/2 * 1/3

1/5

{A}

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Last edited by theCodeToGMAT on Wed May 28, 2014 1:48 am; edited 1 time in total

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Post Sun May 25, 2014 4:31 pm
bml1105 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

OA: A
Number of ways to choose 4 socks from 6 options = 6C4 = (6*5*4*3)/(4*3*2*1) = 15.
Number of ways to choose 2 matching pairs = 3. (RRBB, RRGG, BBGG)
P(selecting 2 matching pairs) = 3/15 = 1/5.

The correct answer is A.

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Last edited by GMATGuruNY on Tue May 27, 2014 2:55 am; edited 1 time in total

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