Pairs of shoes

[Pamirsau]

The question:
The student John has 10 different pairs of shoes. He randomly takes 3 shoes.
1) Calculate how many possibilities does John have.
2) Calculate the probability that from the 3 taken shoes, 2 shoes are from the same pair.

So the first one is easy, it's (20*19*18)/3! = 1140, but I can't figure out the second one. According to the teacher, the answer is 15/38 but I have a feeling that it might be wrong.

 

[Steven G]

What is the probability that none of the 3 shoes make a pairs?

 

[pka]

Of the \(1140\) selections of three shoes how many consist of one pair and one other shoe?

 

[Cubist]

(20*19*18)/3! = 1140

I don't think this is correct. If you ignore the left and right handedness of the shoes, then there are 10*9*8 ways of selecting three distinct shoe types, plus 10*9 ways of selecting a matching pair plus another shoe and this can be done in 3 ways (3 choose 2). So there would be a total of 10*9*8 + 10*9*3 = 990 selections?

 

[pka]

I don't think this is correct. If you ignore the left and right handedness of the shoes, then there are 10*9*8 ways of selecting three distinct shoe types, plus 10*9 ways of selecting a matching pair plus another shoe and this can be done in 3 ways (3 choose 2). So there would be a total of 10*9*8 + 10*9*3 = 990 selections?

\(\bf 1140\) is correct. The whole point is that each pair of shoes contributes two distinct shoes, one left & one right. There is a famous very snide comment by Lord B Russell about such a collection and the axiom of choice.

 

[Pamirsau]

What is the probability that none of the 3 shoes make a pairs?

I get that the probability that none of the 3 shoes make a pair is 16/19. I get the same answer solving in different ways. The first one is (20*18*16)/3! / 1140 = 16/19 meaning we choose the first shoe from 20, the second from 18, so that it doesn't make a pair with the first one and the third one from 16, so that it doesn't make a pair with both the first one and the second one. Then we divide it by the number all possible groups. The second way I did was multiplying the possibilities that the taken shoe doesn't make a pair with shoes taken before it. 1*18/19*16/18 = 16/19 meaning that the first shoe has a 100% chance to not make a pair because it's the first one, then the second one has the chance of 18/19 because there are 19 shoes left and only 18 wouldn't make a pair and the third one has the chance of 16/18 because there are 18 shoes left of which 16 wouldn't make a pair. I suppose if I got the same answer solving in two different ways, it's correct? So the probability that three taken shoes have a pair is 1-16/19=3/19. Is this right?

 

[Cubist]

After re-reading the question, my post #4 is wrong. The question would say if we had to disregard whether the shoe was left or right handed.

 

[Steven G]

You have 10 pairs of shoes and you want to pick 3 where there is no pair.

So from the 10 you choose 3 in ₁₀C₃ways. Now from each pair you want to choose 1. This can be done in 2^3 = 8 ways. Therefore the number of ways to not get a pair is 8*₁₀C₃ = 960 ways and 960/1140 = 16/19