What Is the Probability of Getting at Least One Pair of Shoes from Five Pairs?

[Alexsandro]

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- {5 \choose 0.5 \choose 4 + 5 \choose 1.4 \choose 3 + 5 \choose 2.3 \choose 2 + 5 \choose 3.2 \choose 1 + 5 \choose 4.1 \choose 0}/10 \choose 4.

 

[gjt01]

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?

 

[Hurkyl]

Don't multiple post.

 

[HallsofIvy]

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.

 

[AKG]

I don't know how to reach that answer either. I especially don't know how to reach an answer that has 0 \choose 1 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)/{10}\choose 4
= 1 - 5 \choose 42 \choose 1/210
= 1 - 1/21
= 20/21

 

[Alexsandro]

I don't know how to reach that answer either. I especially don't know how to reach an answer that has 0 \choose 1 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)/{10}\choose 4
= 1 - 5 \choose 42 \choose 1/210
= 1 - 1/21
= 20/21

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

I wrote wrong, not is 0 \choose 1, the correct is 1 \choose 0. I repaired it.

 

[Mithal]

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