Question about pairs of shoes

  • Thread starter Alexsandro
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  • #1
Alexsandro
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I tryed to do this question of many ways, but I couldn't reach the answer. Could someone help me ?

"Four shoes are taken at random from five differents pairs. What is the probability that there is at least one pair among them" ?

The answer to this question is below, but I don't know how I reach it:


1- {[itex]5 \choose 0[/itex].[itex]5 \choose 4[/itex] + [itex]5 \choose 1[/itex].[itex]4 \choose 3[/itex] + [itex]5 \choose 2[/itex].[itex]3 \choose 2[/itex] + [itex]5 \choose 3[/itex].[itex]2 \choose 1[/itex] + [itex]5 \choose 4[/itex].[itex]1 \choose 0[/itex]}/[itex]10 \choose 4[/itex].
 
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Answers and Replies

  • #2
gjt01
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if there are 10 shoes, and you take away 4, logically you would always have at least one pair left, but no more then three. Or am I understanding your question entirly wrong?
 
  • #3
Hurkyl
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Don't multiple post. :grumpy:
 
  • #4
HallsofIvy
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gjt01 said:
if there are 10 shoes, and you take away 4, logically you would always have at least one pair left, but no more then three. Or am I understanding your question entirly wrong?

Yes, you are misunderstanding. He wants a pair in the shoes "taken", not in the shoes left.
 
  • #5
AKG
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I don't know how to reach that answer either. I especially don't know how to reach an answer that has [itex]0 \choose 1[/itex] in it. The answer I got was:

P(at least 1 pair)
= 1 - P(no pairs)
= 1 - (# of ways to select 4 shoes with no pairs)/(# of ways to select 4 shoes)
= 1 - (# of ways to choose 4 pairs to select from * # of ways to pick one shoe from a given pair)/[itex]{10}\choose 4[/itex]
= 1 - [itex]5 \choose 4[/itex][itex]2 \choose 1[/itex]/210
= 1 - 1/21
= 20/21
 
  • #6
Alexsandro
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AKG said:
I don't know how to reach that answer either. I especially don't know how to reach an answer that has [itex]0 \choose 1[/itex] in it. The answer I got was:

P(at least 1 pair)
= 1 - P(no pairs)
= 1 - (# of ways to select 4 shoes with no pairs)/(# of ways to select 4 shoes)
= 1 - (# of ways to choose 4 pairs to select from * # of ways to pick one shoe from a given pair)/[itex]{10}\choose 4[/itex]
= 1 - [itex]5 \choose 4[/itex][itex]2 \choose 1[/itex]/210
= 1 - 1/21
= 20/21

------------------------

I wrote wrong, not is [itex]0 \choose 1[/itex], the correct is [itex]1 \choose 0[/itex]. I repaired it.
 
  • #7
Mithal
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I found simpler way with the same answer as yours

From five pairs of shoes , choosing four pairs is 5 c 4

within the four pairs of shoes , we have 2 ^ 4 possibilities

Hence the number of combinations with no pairs chosen is =

(2^4) * 5 c 4 = 80

p ( at least one pair is chosen) = 1- (80)/ 10 c 4
 

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