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In packing for a trip, Sarah puts three pairs of socks - one
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Updated on: 28 Jun 2017, 19:19
Updated on: 28 Jun 2017, 19:19
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Question Stats:
63% (02:19) correct
36% (01:30) wrong based on 22 sessions
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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
(A) 1/5
(B) 1/4
(C) 1/3
(D) 2/3
(E) 4/5
Re: In packing for a trip, Sarah puts three pairs of socks - one
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20 Jun 2017, 14:23
20 Jun 2017, 14:23
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Explanation
Let us assume all the socks are distinct! I.E. red1, red2, blue1, blue2, green1 and green2.
Let us assume you pull out a red sock its probability is \(\frac{2}{6}\).
next red sock is \(\frac{1}{5}\).
Next you pull blue \(\frac{2}{4}\).
and the other blue \(\frac{1}{3}\).
Now we have the probability of the event (red1, red2, blue1, blue2 ) = \(\frac{2}{6}\frac{1}{5}\frac{2}{4}\frac{1}{3}\) = \(\frac{1}{90}\)
But wait this is not the final probability you need not worry about the order so they can be in any order.
To arrange 4 things we have \(4!\) ways.
New probality = \(4! \times \frac{1}{90}\).
Now we can also pull out green sock so we need to choose 2 coors among 3 can be done by \(3C2\) ways or 3 ways.
Final probability = \(4! \times 3 \times \frac{1}{90} = \frac{24}{30} = \frac{4}{5}\).
Please share the source of these question!
I have a strong feeling this is not a GRE question.
Let us assume all the socks are distinct! I.E. red1, red2, blue1, blue2, green1 and green2.
Let us assume you pull out a red sock its probability is \(\frac{2}{6}\).
next red sock is \(\frac{1}{5}\).
Next you pull blue \(\frac{2}{4}\).
and the other blue \(\frac{1}{3}\).
Now we have the probability of the event (red1, red2, blue1, blue2 ) = \(\frac{2}{6}\frac{1}{5}\frac{2}{4}\frac{1}{3}\) = \(\frac{1}{90}\)
But wait this is not the final probability you need not worry about the order so they can be in any order.
To arrange 4 things we have \(4!\) ways.
New probality = \(4! \times \frac{1}{90}\).
Now we can also pull out green sock so we need to choose 2 coors among 3 can be done by \(3C2\) ways or 3 ways.
Final probability = \(4! \times 3 \times \frac{1}{90} = \frac{24}{30} = \frac{4}{5}\).
Please share the source of these question!
I have a strong feeling this is not a GRE question.
Re: In packing for a trip, Sarah puts three pairs of socks - one
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28 Jun 2017, 06:16
28 Jun 2017, 06:16
2
There are three different colour socks, so either it could RRBB+RRGG+BBGG,
Let's find for one case, we can finally multiply with three,
So let’s find for RRBB
= (2/6)*(1/5)*(2/4)*(1/3) * 4!/(2!*2!) = 1/15
4!/(2!*2!) is multiplied as the four socks can pulled out in any order (RRBB,BBRR,RBRB,BRBR,BRRB,RBBR)
There are three different cases possible(as shown above),
So 1/15 * 3 = 1/5
So the answer is A.
Let's find for one case, we can finally multiply with three,
So let’s find for RRBB
= (2/6)*(1/5)*(2/4)*(1/3) * 4!/(2!*2!) = 1/15
4!/(2!*2!) is multiplied as the four socks can pulled out in any order (RRBB,BBRR,RBRB,BRBR,BRRB,RBBR)
There are three different cases possible(as shown above),
So 1/15 * 3 = 1/5
So the answer is A.
