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

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nonameee
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There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yell [#permalink] New post   27 May 2010, 03:41
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54% (01:32) correct 46% (01:50) wrong based on 13 sessions
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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
Show Spoiler
13/21


The book says it's
Show Spoiler
41/42
which is in my opinion a mistake.

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Re: There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yell [#permalink] New post   27 May 2010, 04:59
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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
Show Spoiler
13/21


The book says it's
Show Spoiler
41/42
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)=1-P(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: There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yell [#permalink] New post   08 Mar 2014, 05:02
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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) = 1-8/21= 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: There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yell [#permalink] New post   25 Jul 2014, 03:16
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kiranpradhan wrote:
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!


The correct answer is 13/21, not 41/42. The point is that each out of 4 color socks you are selecting can give either first or second sock. Please read the whole discussion above.
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Re: There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yell [#permalink] New post   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: There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yell [#permalink] New post   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
Show Spoiler
13/21


The book says it's
Show Spoiler
41/42
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)=1-P(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: There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yell [#permalink] New post   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
Show Spoiler
13/21


The book says it's
Show Spoiler
41/42
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)=1-P(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: There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yell [#permalink] New post   29 May 2010, 03:31
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Detran wrote:

wait a sec i dont get it. Why isnt 5 choose 4 \(C^5_4\) ?


It's just another way of writing it \(C^5_4\), \(C^4_5\), \(5C4\).
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Re: There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yell [#permalink] New post   06 Mar 2014, 11:37
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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
Show Spoiler
13/21


The book says it's
Show Spoiler
41/42
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)=1-P(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: There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yell [#permalink] New post   07 Mar 2014, 01:21
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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
Show Spoiler
13/21


The book says it's
Show Spoiler
41/42
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)=1-P(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 [#permalink] New post   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: There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yell [#permalink] New post   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
Show Spoiler
13/21


The book says it's
Show Spoiler
41/42
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)=1-P(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: There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yell [#permalink] New post   03 May 2014, 04:37
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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
Show Spoiler
13/21


The book says it's
Show Spoiler
41/42
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)=1-P(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 [#permalink] New post   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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Re: There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yell [#permalink] New post   23 Jul 2021, 17:11
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Re: There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yell [#permalink]
23 Jul 2021, 17:11
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