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There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yell [#permalink]

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27 May 2010, 02:41

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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?

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?

"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?
_________________

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?

"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.

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?

"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 \(C^5_4\) ?
_________________

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?

"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 ..
_________________

Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !

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?

"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.

Re: There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yell [#permalink]

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08 Mar 2014, 04:02

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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

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?

"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.

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?

"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.
_________________

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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