A hard problem on probability !

Well, I found it hard, maybe it's not that hard for you guys :smile:

There are n pairs of shoes in the stock. Choose randomly 2k shoes from the stock with 2k < n. Find the probability of the case: there is only 1 pair of shoes in these 2k shoes taken.

Reply 1

rubysolstice

I find it hard also. =S You've given n pairs of shoes, and k pairs of shoes are chosen, therefore what's the probability of k being one. I don't think that's enough info.

EDIT: Ignore me. I misunderstood you- sorry!

Reply 2

Jonny W

8

Start by choosing the two shoes from the same pair [n ways].

That leaves (2k - 2) shoes to choose. First choose which pairs these (2k - 2) shoes will come from [(n - 1)C(2k - 2) ways]. Then choose whether each shoe will be a left or right [2^(2k - 2) ways].

So there are n * (n - 1)C(2k - 2) * 2^(2k - 2) ways of choosing 2k shoes with exactly one pair.

Probability = n * (n - 1)C(2k - 2) * 2^(2k - 2) / (2n)C(2k)

Reply 3

Saichu

11

Edit: Gah, i thought it said n shoes, not n pairs.

//Can I assume n is even? I might have to rework this if not (and btw, jony's is probably more well-checked out, i'm just randomizing this for fun).

Total number of ways to choose 2k shoes from n total: 2n C 2k.

There are n pairs of shoes. 2k - 1 must belong to different pairs, so number of ways to choose different pairs: (n) C (2k-1)

Obviously, 2 ways to choose 1 shoe per pair.

Number of ways to choose remaining pair: n - (2k - 1) = n + 1 - 2k

Therefore:

Probability = Number of ways to achieve goal / total number of ways.
Probability = 2*[n C (2k-1)]*(n + 1 - 2k)/(2n C 2k)

Don't ask.. i'm not testing that :biggrin:

Reply 4

Jonny W

8

Saichu

Edit: Gah, i thought it said n shoes, not n pairs.

//Can I assume n is even? I might have to rework this if not (and btw, jony's is probably more well-checked out, i'm just randomizing this for fun).

Total number of ways to choose 2k shoes from n total: 2n C 2k.

There are n pairs of shoes. 2k - 1 must belong to different pairs, so number of ways to choose different pairs: (n) C (2k-1)

OK so far. Now we have to choose which of the (2k - 1) pairs we're going to take both shoes from [(2k - 1) ways], and for each of the remaining (2k - 2) pairs whether we're going to take the left or right shoe [2^(2k - 2) ways].

So there are nC(2k - 1) * (2k - 1) * 2^(2k - 2) ways of choosing 2k shoes with exactly one pair.

Probability = nC(2k - 1) * (2k - 1) * 2^(2k - 2) / (2n)C(2k)

Reply 5

Saichu

11

erm, it's one of the n pairs we're gonna take both shoes from (n ways) and that leaves n - 1 total pairs remaining, of which we choose 2k - 1 pairs to take one shoe from. The number of ways to choose 2k - 1 pairs out of n-1 is (n-1)C(2k-1). Mm, but yeah.. two ways to choose from Each pair, so that's multiplied by 2^(2k-1). So from what i can tell, the answer's a cross between your first and second :biggrin:

Probability = n*(n - 1)C(2k - 1) * 2^(2k - 1) / (2n)C(2k)

(Someone really needs to check this)

---

ergh, or doing it my first way... choose the one per pair ones first. that means you have nC(2k - 1) combinations of pairs.. and 2 shoes per pair... so 2^(2k-1) multiplied to that for number of ways. Next, n-(2k-1) = n + 1 - 2k remaining options for the whole pair, so...

Probability = nC(2k - 1)*2^(2k-1)*(n + 1 - 2k )/(2n)C(2k)

Reply 6

Jonny W

8

Saichu

erm, it's one of the n pairs we're gonna take both shoes from (n ways) and that leaves n - 1 total pairs remaining, of which we choose 2k - 1 pairs to take one shoe from. The number of ways to choose 2k - 1 pairs out of n-1 is (n-1)C(2k-1). Mm, but yeah.. two ways to choose from Each pair, so that's multiplied by 2^(2k-1). So from what i can tell, the answer's a cross between your first and second :biggrin:

Probability = n*(n - 1)C(2k - 1) * 2^(2k - 1) / (2n)C(2k)

After we've picked the complete pair, we need to choose another 2k - 2 shoes (not 2k - 1). So it should be "(n - 1)C(2k - 2) * 2^(2k - 2)" not "(n - 1)C(2k - 1) * 2^(2k - 1)".

My first and second answers are equivalent because

n * (n - 1)C(2k - 2)
= n * (n - 1)! / [(2k - 2)!(n + 1 - 2k)!]
= n! / [(2k - 2)!(n + 1 - 2k)!]
= (2k - 1) * n! / [(2k - 1)!(n + 1 - 2k)!]
= (2k - 1) * nC(2k - 1)

Reply 7

Saichu

11

Ah, yes. 2 for a pair. Thank you.

Reply 8

BCHL85

OP

12

Thank you all. The answer is correct :smile:

A hard problem on probability !

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