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There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yell
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27 May 2010, 03:41
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There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yellow, 2 green. You select randomly four socks together. What is the probability that you'll get at least two of the same color? I think it's The book says it's which is in my opinion a mistake.



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Re: Probability question: 10 socks
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27 May 2010, 04:59
nonameee wrote: There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yellow, 2 green. You select randomly four socks together. What is the probability that you'll get at least two of the same color? I think it's The book says it's which is in my opinion a mistake. I think you are right. "Probability that you'll get at least two of the same color"  means at least one pair (one pair or two pairs out of 4 socks). P(at least one pair)=1P(no pair) > no pair means all socks must be of different colors > \(P=1\frac{C^4_5*2^4}{C^4_{10}}=\frac{13}{21}\). \(C^4_5\)  # of ways to choose 4 different colors out of 5, basically # of ways to choose which 4 color socks will give us one sock (in this case we obviously will have 4 different colors); \(2^4\)  each of 4 chosen colors can give left or right sock; \(C^4_{10}\)  total # of ways to choose 4 socks out of 10. What solution was given in the book for 41/42?
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Re: Probability question: 10 socks
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27 May 2010, 05:02
Quote: What solution was given in the book for 41/42? I think they've forgotten to multiply by 2^4. But the solution was the same you gave.



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Re: Probability question: 10 socks
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28 May 2010, 21:26
Thanks for the explanation. Bunuel wrote: nonameee wrote: There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yellow, 2 green. You select randomly four socks together. What is the probability that you'll get at least two of the same color? I think it's The book says it's which is in my opinion a mistake. I think you are right. "Probability that you'll get at least two of the same color"  means at least one pair (one pair or two pairs out of 4 socks). P(at least one pair)=1P(no pair) > no pair means all socks must be of different colors > \(P=1\frac{C^4_5*2^4}{C^4_10}=\frac{13}{21}\). \(C^4_5\)  # of ways to choose 4 different colors out of 5, basically # of ways to choose which 4 color socks will give us one sock (in this case we obviously will have 4 different colors); \(2^4\)  each of 4 chosen colors can give left or right sock; \(C^4_{10}\)  total # of ways to choose 4 socks out of 10. What solution was given in the book for 41/42?



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Re: Probability question: 10 socks
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28 May 2010, 22:07
Bunuel wrote: nonameee wrote: There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yellow, 2 green. You select randomly four socks together. What is the probability that you'll get at least two of the same color? I think it's The book says it's which is in my opinion a mistake. I think you are right. "Probability that you'll get at least two of the same color"  means at least one pair (one pair or two pairs out of 4 socks). P(at least one pair)=1P(no pair) > no pair means all socks must be of different colors > \(P=1\frac{C^4_5*2^4}{C^4_10}=\frac{13}{21}\). \(C^4_5\)  # of ways to choose 4 different colors out of 5, basically # of ways to choose which 4 color socks will give us one sock (in this case we obviously will have 4 different colors); \(2^4\)  each of 4 chosen colors can give left or right sock; \(C^4_{10}\)  total # of ways to choose 4 socks out of 10. What solution was given in the book for 41/42? wait a sec i dont get it. Why isnt 5 choose 4 \(C^5_4\) ?
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Re: Probability question: 10 socks
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29 May 2010, 03:31



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Re: Probability question: 10 socks
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06 Mar 2014, 11:37
Bunuel wrote: nonameee wrote: There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yellow, 2 green. You select randomly four socks together. What is the probability that you'll get at least two of the same color? I think it's The book says it's which is in my opinion a mistake. I think you are right. "Probability that you'll get at least two of the same color"  means at least one pair (one pair or two pairs out of 4 socks). P(at least one pair)=1P(no pair) > no pair means all socks must be of different colors > \(P=1\frac{C^4_5*2^4}{C^4_10}=\frac{13}{21}\). \(C^4_5\)  # of ways to choose 4 different colors out of 5, basically # of ways to choose which 4 color socks will give us one sock (in this case we obviously will have 4 different colors); \(2^4\)  each of 4 chosen colors can give left or right sock; \(C^4_{10}\)  total # of ways to choose 4 socks out of 10. What solution was given in the book for 41/42? May i knw the logic behind this? m not getting the numerator part ..
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Re: Probability question: 10 socks
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07 Mar 2014, 01:21
sanjoo wrote: Bunuel wrote: nonameee wrote: There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yellow, 2 green. You select randomly four socks together. What is the probability that you'll get at least two of the same color? I think it's The book says it's which is in my opinion a mistake. I think you are right. "Probability that you'll get at least two of the same color"  means at least one pair (one pair or two pairs out of 4 socks). P(at least one pair)=1P(no pair) > no pair means all socks must be of different colors > \(P=1\frac{C^4_5*2^4}{C^4_10}=\frac{13}{21}\). \(C^4_5\)  # of ways to choose 4 different colors out of 5, basically # of ways to choose which 4 color socks will give us one sock (in this case we obviously will have 4 different colors); \(2^4\)  each of 4 chosen colors can give left or right sock;\(C^4_{10}\)  total # of ways to choose 4 socks out of 10. What solution was given in the book for 41/42? May i knw the logic behind this? m not getting the numerator part .. If 4 out of 5 pairs of socks will give one sock, we won't have a matching pair, we'll have 4 different color socks and this is exactly what we are counting in the numerator. Hope it's clear.
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Re: There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yell
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08 Mar 2014, 05:02
I solved this one be listing method
Prob of at least getting a pair = 1 prob(all different at four attempts) = 1( 1 x 8/9 x 6/8 x 4/7) = 18/21= 13/21
a) 1 st attempt = getting any color =1 b) 2nd attempt = not getting the color picked in a). = 8/9 c) 3rd attempt = not getting the two colors above = 6/8 d) not getting any of the four colors above = 4/7



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Re: Probability question: 10 socks
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12 Mar 2014, 11:52
Dear Bunuel
I did 10c1X1c1 X 8c1X6c1/ 10c4 + 10c1X1c1 X 8c1X1c1/10c4. Answer is wrong. Please help!
Thanks & regards



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Re: Probability question: 10 socks
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03 May 2014, 02:00
Why we took 2^4 ? what if we want to solve it by straight forward method ? that is considering P ( getting two socks atleast of same color ) ? Bunuel wrote: nonameee wrote: There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yellow, 2 green. You select randomly four socks together. What is the probability that you'll get at least two of the same color? I think it's The book says it's which is in my opinion a mistake. I think you are right. "Probability that you'll get at least two of the same color"  means at least one pair (one pair or two pairs out of 4 socks). P(at least one pair)=1P(no pair) > no pair means all socks must be of different colors > \(P=1\frac{C^4_5*2^4}{C^4_{10}}=\frac{13}{21}\). \(C^4_5\)  # of ways to choose 4 different colors out of 5, basically # of ways to choose which 4 color socks will give us one sock (in this case we obviously will have 4 different colors); \(2^4\)  each of 4 chosen colors can give left or right sock; \(C^4_{10}\)  total # of ways to choose 4 socks out of 10. What solution was given in the book for 41/42?



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Re: Probability question: 10 socks
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03 May 2014, 04:37
sunny3011 wrote: Why we took 2^4 ? what if we want to solve it by straight forward method ? that is considering P ( getting two socks atleast of same color ) ? Bunuel wrote: nonameee wrote: There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yellow, 2 green. You select randomly four socks together. What is the probability that you'll get at least two of the same color? I think it's The book says it's which is in my opinion a mistake. I think you are right. "Probability that you'll get at least two of the same color"  means at least one pair (one pair or two pairs out of 4 socks). P(at least one pair)=1P(no pair) > no pair means all socks must be of different colors > \(P=1\frac{C^4_5*2^4}{C^4_{10}}=\frac{13}{21}\). \(C^4_5\)  # of ways to choose 4 different colors out of 5, basically # of ways to choose which 4 color socks will give us one sock (in this case we obviously will have 4 different colors); \(2^4\)  each of 4 chosen colors can give left or right sock; \(C^4_{10}\)  total # of ways to choose 4 socks out of 10. What solution was given in the book for 41/42? Can you please tell me which part of below explanation is not clear? \(C^4_5\)  # of ways to choose 4 different colors out of 5, basically # of ways to choose which 4 color socks will give us one sock (in this case we obviously will have 4 different colors); \(2^4\)  each of 4 chosen colors can give left or right sock. As for direct approach: it would be lengthier way to get the answer. You should count two cases: A. one par of matching socks with two socks of different color; B. two pairs of matching socks.
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Re: There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yell
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24 Jul 2014, 18:07
Total Number of Possible Combination= 10C4= 210
Total number of Combination where socks is not of same colour= 5C4= 5
Probability of Same= 1 5/210= 41/42
PS: In socks there is nothing as left & right. So the factor of 2^4 is not required.
Hope it is clear & book is right!



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There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yell
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