Oct 15 12:00 PM PDT  01:00 PM PDT Join this live GMAT class with GMAT Ninja to learn to conquer your fears of long, kooky GMAT questions. Oct 16 08:00 PM PDT  09:00 PM PDT EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth $100 with the 3 Month Pack ($299) Oct 19 07:00 AM PDT  09:00 AM PDT Does GMAT RC seem like an uphill battle? eGMAT is conducting a free webinar to help you learn reading strategies that can enable you to solve 700+ level RC questions with at least 90% accuracy in less than 10 days. Sat., Oct 19th at 7 am PDT Oct 20 07:00 AM PDT  09:00 AM PDT Get personalized insights on how to achieve your Target Quant Score.
Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 26 Feb 2018
Posts: 76
Location: United Arab Emirates
GMAT 1: 710 Q47 V41 GMAT 2: 770 Q49 V47

Re: M0206
[#permalink]
Show Tags
10 Mar 2018, 15:05
Method for this question
When ever you see "at least" this is a hint for using the formula P (at least) = 1  (P less than) (which is easier than considering all the individual scenarios for 2, 3 or 4 matching socks and summing them. That would be very time consuming as we could have WWWW, WWWB, WWWG, WWBB, WWGG WWBG BBWG BBGG BBBW BBBG GGBW etc)
So what is the probability of NOT picking at least 2 socks of the same colour? IE less than 2 matching socks? Ie. NO matching socks. For that we need all 4 socks of a different colour. However, we only have 3 colours, so the probability of this is zero. 10=1



Manager
Joined: 18 Jul 2018
Posts: 52
Location: United Arab Emirates

Re: M0206
[#permalink]
Show Tags
31 Mar 2019, 01:27
Bunuel wrote: Official Solution:
There are 5 pairs of white, 3 pairs of black and 2 pairs of grey socks in a drawer. If four individual socks are picked at random what is the probability of getting at least two socks of the same color?
A. \(\frac{1}{5}\) B. \(\frac{2}{5}\) C. \(1\) D. \(\frac{3}{5}\) E. \(\frac{4}{5}\)
No formula is need to answer this question. The trick here is that we have only 3 different color socks but we pick 4 socks, which ensures that in ANY case we'll have at least one pair of the same color (if 3 socks we pick are of the different color, then the 4th sock must match with either of previously picked one). P=1.
Answer: C Hi XylanI understand logic works perfectly here! What about complement rule?? THANKS IN ADVANCE



Manager
Status: The darker the night, the nearer the dawn!
Joined: 16 Jun 2018
Posts: 179

JIAA wrote: Bunuel wrote: Official Solution:
There are 5 pairs of white, 3 pairs of black and 2 pairs of grey socks in a drawer. If four individual socks are picked at random what is the probability of getting at least two socks of the same color?
A. \(\frac{1}{5}\) B. \(\frac{2}{5}\) C. \(1\) D. \(\frac{3}{5}\) E. \(\frac{4}{5}\)
No formula is need to answer this question. The trick here is that we have only 3 different color socks but we pick 4 socks, which ensures that in ANY case, we'll have at least one pair of the same color (if 3 socks we pick are of the different color, then the 4th sock must match with either of previously picked one). P=1.
Answer: C Hi XylanI understand logic works perfectly here! What about complement rule?? THANKS IN ADVANCE Given Data:10 White  6 Black  4 Grey JIAA By the complement rule: Total = Case1 (when probability is at least 2) + Case2 (When probability is less than 2)To purposefully find the probability for which we get less than 2 socks of the same color, Let's fill the 3 of the 4 empty spaces: W  B  G  __ 4thspace __ As we can see, the 4thspace can ONLY be filled by any of the already existing colors of the socks , i.e., by White, Black OR Grey, thus NO scenario is possible that can fit the conditon. Hence, the probability for which we get less than 2 socks of the same color = \(0\)  The Complementing Case Therefore, ALL the possible cases have at least 2 socks of the same colors, i.e., Probability = 1
_________________
 “The trouble is, you think you have time.” – Buddha Giving Kudos is the best way to encourage and appreciate people.



Manager
Joined: 18 Jul 2018
Posts: 52
Location: United Arab Emirates

Re: M0206
[#permalink]
Show Tags
01 Apr 2019, 06:18
Xylan wrote: JIAA wrote: Bunuel wrote: Official Solution:
There are 5 pairs of white, 3 pairs of black and 2 pairs of grey socks in a drawer. If four individual socks are picked at random what is the probability of getting at least two socks of the same color?
A. \(\frac{1}{5}\) B. \(\frac{2}{5}\) C. \(1\) D. \(\frac{3}{5}\) E. \(\frac{4}{5}\)
No formula is need to answer this question. The trick here is that we have only 3 different color socks but we pick 4 socks, which ensures that in ANY case, we'll have at least one pair of the same color (if 3 socks we pick are of the different color, then the 4th sock must match with either of previously picked one). P=1.
Answer: C Hi XylanI understand logic works perfectly here! What about complement rule?? THANKS IN ADVANCE Given Data:10 White  6 Black  4 Grey JIAA By the complement rule: Total = Case1 (when probability is at least 2) + Case2 (When probability is less than 2)To purposefully find the probability for which we get less than 2 socks of the same color, Let's fill the 3 of the 4 empty spaces: W  B  G  __ 4thspace __ As we can see, the 4thspace can ONLY be filled by any of the already existing colors of the socks , i.e., by White, Black OR Grey, thus NO scenario is possible that can fit the conditon. Hence, the probability for which we get less than 2 socks of the same color = \(0\)  The Complementing Case Therefore, ALL the possible cases have at least 2 socks of the same colors, i.e., Probability = 1THANK YOU XylanMUCH APPRECIATED !!!



VP
Joined: 14 Feb 2017
Posts: 1170
Location: Australia
Concentration: Technology, Strategy
GMAT 1: 560 Q41 V26 GMAT 2: 550 Q43 V23 GMAT 3: 650 Q47 V33 GMAT 4: 650 Q44 V36
WE: Management Consulting (Consulting)

Re: M0206
[#permalink]
Show Tags
11 Jun 2019, 03:00
I got this wrong, but I didn't think it was this simple when reviewing the explanation. I think a lot of people are over complicating this. There are 10 White, 6 Blue and 4 Grey. What is the probability that I pick a matching pair of socks if I am to choose 4 individual socks? 4 Choices: _ _ _ _ Randomize this: W B G ? ? = we must choose either White, Blue or Grey. So we have a 100% chance of picking a matching pair we choose W B G (W/B/G) = Matching pair.
_________________
Goal: Q49, V41
+1 Kudos if I have helped you



Intern
Joined: 12 Nov 2016
Posts: 7

Re: M0206
[#permalink]
Show Tags
11 Jun 2019, 05:58
chetan2u wrote: MEOWSER wrote: Bunuel,
Thanks for the question. If it were phrased differently and we only picked 3 individual socks instead of 4 individual socks, how would we approach the question? the recommended way would be.. in this case you will have to find the probability of all three being different... since there are three different socks , the ways to pick up three amongst th three would be 3!=6... now the probability in each way=\(\frac{10}{20}*\frac{6}{19}*\frac{4}{18}\)=\(\frac{2}{57}\)... so for 6 ways=6*\(\frac{2}{57}\)= \(\frac{12}{57}\).. so the prob of not having all different or in other words having atleast two of same colour= 1\(\frac{12}{57}\)=\(\frac{45}{57}\) Hello Please explain how does the selection order (and thereby an extra factor of 3!) matter in the above solution provided if we were to make a selection of 3 individual socks. Thanks







Go to page
Previous
1 2
[ 26 posts ]



