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In packing for a trip, Sarah puts three pairs of socks  one [#permalink]
20 Jun 2017, 11:09
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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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Last edited by pranab01 on 28 Jun 2017, 19:19, edited 2 times in total.




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Re: In packing for a trip, Sarah puts three pairs of socks  one [#permalink]
20 Jun 2017, 14:23
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ExplanationLet 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.
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Re: In packing for a trip, Sarah puts three pairs of socks  one [#permalink]
28 Jun 2017, 06:16
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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.



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Re: In packing for a trip, Sarah puts three pairs of socks  one [#permalink]
26 Jul 2017, 01:15
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sandy wrote: 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 colors 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. Hi Could you reexplain the green sock bit? So out of 4 socks, we first check for 2 pairs (which is 1/90). Then what is the scenario for green socks? Thank you!



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Re: In packing for a trip, Sarah puts three pairs of socks  one [#permalink]
26 Jul 2017, 03:26
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nancyjose wrote: sandy wrote: 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 colors 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. Hi Could you reexplain the green sock bit? So out of 4 socks, we first check for 2 pairs (which is 1/90). Then what is the scenario for green socks? Thank you! Hello nancyjose, try this approach if it makes it easier to understand We have 3 pairs in total i.e 2red, 2 blue and 2 green and 6 number of socks in total. Now Select 2 pairs out of 3 i.e = 3c2 = 3 (since we need exactly 2pair) Selecting 4 socks randomly = 6c4 = 15 Probability = 3/15 = 1/5
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Re: In packing for a trip, Sarah puts three pairs of socks  one [#permalink]
15 Dec 2017, 07:01
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pranab01 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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