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5 pairs of socks are randomly distributed in a bag
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25 Aug 2018, 22:52
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67% (00:41) correct 33% (00:47) wrong based on 101 sessions
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5 pairs of socks are randomly distributed in a bag.If you reach in and select 2 socks, what is the probability that you are holding a matched pair? a) 1/100 b) 1/90 c) 1/10 d) 1/9 e) 1/5
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Re: 5 pairs of socks are randomly distributed in a bag
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25 Aug 2018, 23:36
GMATbuster92 wrote: 5 pairs of socks are randomly distributed in a bag.If you reach in and select 2 socks, what is the probability that you are holding a matched pair? a) 1/100 b) 1/90 c) 1/10 d) 1/9 e) 1/5 Sample space= No of ways of selecting 2 nos. of socks out of 10 nos. of socks=10C2 Favourable ways= No of ways selecting 1 pair of socks out of 5 pairs=5C1 So, required probability=\(\frac{5C1}{10C2}\)=\(\frac{5}{5*9}\)=\(\frac{1}{9}\) Ans. (D)
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Re: 5 pairs of socks are randomly distributed in a bag
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27 Aug 2018, 20:05
PKN wrote: GMATbuster92 wrote: 5 pairs of socks are randomly distributed in a bag.If you reach in and select 2 socks, what is the probability that you are holding a matched pair? a) 1/100 b) 1/90 c) 1/10 d) 1/9 e) 1/5 Sample space= No of ways of selecting 2 nos. of socks out of 10 nos. of socks=10C2Favourable ways= No of ways selecting 1 pair of socks out of 5 pairs=5C1So, required probability=\(\frac{5C1}{10C2}\)=\(\frac{5}{5*9}\)=\(\frac{1}{9}\) Ans. (D) Hi, Are you not comparing apples against the oranges? I mean, blue above says socks and green says pairs? I used this technique, please correct me if I am wrong. Base: 10C2 5 Pairs>10 Socks and 2 Selections  Makes Sense Fav Outcomes: 2C2*5 [1 Option, 5 Pairs] Prob 5/45=1/9 "D" Thanks, Vikram



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Re: 5 pairs of socks are randomly distributed in a bag
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27 Aug 2018, 21:20
vikgupta07 wrote: PKN wrote: GMATbuster92 wrote: 5 pairs of socks are randomly distributed in a bag.If you reach in and select 2 socks, what is the probability that you are holding a matched pair? a) 1/100 b) 1/90 c) 1/10 d) 1/9 e) 1/5 Sample space= No of ways of selecting 2 nos. of socks out of 10 nos. of socks=10C2Favourable ways= No of ways selecting 1 pair of socks out of 5 pairs=5C1So, required probability=\(\frac{5C1}{10C2}\)=\(\frac{5}{5*9}\)=\(\frac{1}{9}\) Ans. (D) Hi, Are you not comparing apples against the oranges? I mean, blue above says socks and green says pairs? I used this technique, please correct me if I am wrong. Base: 10C2 5 Pairs>10 Socks and 2 Selections  Makes Sense Fav Outcomes: 2C2*5 [1 Option, 5 Pairs] Prob 5/45=1/9 "D" Thanks, Vikram Hi vikgupta07 . aa bb cc dd ee It's not about \(\frac{#pairs}{#socks}\) or \(\frac{#socks}{#socks}\), rather it is \(\frac{#pairs}{#pairs}\), where Numerator=favorable events(in pairs)= No of times or ways we can hold one matching pair of socks out of 5 pairs=5C1 (aa,bb,cc,dd,ee) Denominator=Total no of events(in pairs)=No of times or ways we can hold a pair of socks(selection of 2nos) out of 10 socks=10C2 (The outcomes of 10C2 are in pair. Such as ab, ac, aa,bb,bc,bd etc [Matched pair+Unmatched pair]) Hope its clear. P.S: Not sure how you have arrived at 2C2*5.
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Re: 5 pairs of socks are randomly distributed in a bag
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27 Aug 2018, 23:54
Thanks for the reply.
For the numerator part, choosing two correct socks out of a pair: 2C2
The choice will be a ‘equally likely’ criterion for all the 5 pairs.
2C2+2C2+2C2+2C2+2C2
And hence 5*2C2.
Still working on my probability concepts. Thanks for bearing with me ?
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5 pairs of socks are randomly distributed in a bag
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28 Aug 2018, 00:22
GMATbuster92 wrote: 5 pairs of socks are randomly distributed in a bag.If you reach in and select 2 socks, what is the probability that you are holding a matched pair? a) 1/100 b) 1/90 c) 1/10 d) 1/9 e) 1/5 1st sock can be selected in 10 ways 2nd one can be selected in only one way p(1st sock) = 10/10 p(2nd sock) = 1/9 p(of selecting 2 of same colour) = 10/90 = 1/9



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5 pairs of socks are randomly distributed in a bag
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28 Aug 2018, 00:26
GMATbuster92 wrote: 5 pairs of socks are randomly distributed in a bag.If you reach in and select 2 socks, what is the probability that you are holding a matched pair? a) 1/100 b) 1/90 c) 1/10 d) 1/9 e) 1/5 here is my approach guys We have following pair of socks: AA, BB, CC, DD, EE hence we have FIVE possibilities to choose a a matching pair of sock Total number of socks 10 Total number of possibilities to choose any two socks from 10  > \(\frac{10!}{2!(102)!} = 45\) \(\frac{5}{45}\) = \(\frac{1}{9}\)



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Re: 5 pairs of socks are randomly distributed in a bag
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10 Sep 2018, 05:52
Hi stnePlease find below another way similar to the one I have posted PKN has already given an appropriate explanation , I just wanted to show another way to solve the question as per the question we have 5 pairs consider below 1122334455 are the given pairs no of ways we can select one pair = 5c1 = 5 ways total number of ways = 10c2=45 probability = 5/45 = 1/9 I have solved this question the same way as I have done below https://gmatclub.com/forum/jessicasla ... s#p2128841



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Re: 5 pairs of socks are randomly distributed in a bag
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10 Sep 2018, 06:00
CounterSniper wrote: Hi stnePlease find below another way similar to the one I have posted PKN has already given an appropriate explanation , I just wanted to show another way to solve the question as per the question we have 5 pairs consider below 1122334455 are the given pairs no of ways we can select one pair = 5c1 = 5 ways total number of ways = 10c2=45 probability = 5/45 = 1/9 I have solved this question the same way as I have done below https://gmatclub.com/forum/jessicasla ... s#p2128841Actually this seems different from the other post. Ways of selecting the first sock is 10C1 for the second sock there is one one way hence 10C1 *1 Total ways of selecting 2 socks from 10 socks is 10C2 = 45 hence probability is 10/45 = 2/9 https://gmatclub.com/forum/jessicasla ... l#p2128841
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Re: 5 pairs of socks are randomly distributed in a bag
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10 Sep 2018, 06:05
stne wrote: CounterSniper wrote: Hi stnePlease find below another way similar to the one I have posted PKN has already given an appropriate explanation , I just wanted to show another way to solve the question as per the question we have 5 pairs consider below 1122334455 are the given pairs no of ways we can select one pair = 5c1 = 5 ways total number of ways = 10c2=45 probability = 5/45 = 1/9 I have solved this question the same way as I have done below https://gmatclub.com/forum/jessicasla ... s#p2128841Actually this seems different from the other post. Ways of selecting the first sock is 10C1 for the second sock there is one one way hence 10C1 *1 Total ways of selecting 2 socks from 10 socks is 10C2 = 45 hence probability is 10/45 = 2/9 https://gmatclub.com/forum/jessicasla ... l#p2128841when we select the first sock , only 9 are left !! so the remaining one can be selected in 9c1=9 ways so probability  1/9



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Re: 5 pairs of socks are randomly distributed in a bag
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10 Sep 2018, 06:07
CounterSniper wrote: stne wrote: CounterSniper wrote: Hi stnePlease find below another way similar to the one I have posted PKN has already given an appropriate explanation , I just wanted to show another way to solve the question as per the question we have 5 pairs consider below 1122334455 are the given pairs no of ways we can select one pair = 5c1 = 5 ways total number of ways = 10c2=45 probability = 5/45 = 1/9 I have solved this question the same way as I have done below https://gmatclub.com/forum/jessicasla ... s#p2128841Actually this seems different from the other post. Ways of selecting the first sock is 10C1 for the second sock there is one one way hence 10C1 *1 Total ways of selecting 2 socks from 10 socks is 10C2 = 45 hence probability is 10/45 = 2/9 https://gmatclub.com/forum/jessicasla ... l#p2128841when we select the first sock , only 9 are left !! so the remaining one can be selected in 9c1=9 ways so probability  1/9 OOps ! My bad , realized my mistake. Thank you.
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Re: 5 pairs of socks are randomly distributed in a bag &nbs
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