Probability & Combination. N pairs of shoes in closet.

Question

7 pairs of shoes (i.e., 14 shoes) are in a closet. 4 persons whose shoes are among the 7 pairs randomly selected 2 shoes each from the closet. What is the probability that exactly 2 persons selected their pairs of shoes?

My attempt:

$1\frac{12}{13}\frac{10}{12}\frac{8}{11}$

is the P(of no pairs)

Another attempt:

$\frac{\:2!}{0!\:\cdot \left(2\:-\:0\right)!}\: \cdot \frac{\:14!}{5!\:\cdot \left(14\:-\:5\right)!}\:$ = Possible outcomes for the ith person selects their own pair of shoes

Answer 1

Number the persons with $1,2,3,4$.

Let $A,B,C,D$ denote the events that persons $1,2,3$ and $4$ respectively picks his own shoes.

Then by symmetry it is clear that the probability that exactly $2$ persons pick their own shoes is: $$\binom{4}{2}P\left(A\cap B\cap C^{\complement}\cap D^{\complement}\right)$$

Here: $$P\left(A\cap B\cap C^{\complement}\cap D^{\complement}\right)=P\left(A\right)P\left(B\mid A\right)P\left(C^{\complement}\cap D^{\complement}\mid A\cap B\right)$$

and: $$P\left(C^{\complement}\cap D^{\complement}\mid A\cap B\right)=1-P\left(C\cup D\mid A\cap B\right)=$$$$1-P\left(C\mid A\cap B\right)-P\left(D\mid A\cap B\right)+P\left(C\cap D\mid A\cap B\right)=$$$$1-2P\left(C\mid A\cap B\right)+P\left(C\mid A\cap B\right)P\left(D\mid A\cap B\cap C\right)$$

---

Can you find $P\left(A\right),P\left(B\mid A\right),P\left(C\mid A\cap B\right)$ and $P\left(D\mid A\cap B\cap C\right)$ yourself?

Answer 2

We have $14$ shoes in the closet and $4$ people who pick $2$ shoes each. Our $\Omega$ can be seen as $\Omega = \{ (x_1,x_2,x_3,x_4 ): x_i \in \mathcal P_2(\{1,...,14\}, x_i \cap x_j = \emptyset$ for $ i \neq j$, and $ i,j \in \{1,...,4\} \} $

$\mathcal P_2(X)$ denote set of subsets $A$ of $X$, such that $|A| = 2$.

$x_i$ is the pair of shoes that $i'$th person pick.

That is, we have $|\Omega| = {14 \choose 2}{12 \choose 2}{10 \choose 2}{8 \choose 2} $

Now, we look at event $E$ - exactly 2 people select their pair of shoes.

We have ${4 \choose 2}$ ways to choose those people, and for every choice the rest is symetrical, so let's figure out the answer for case: $1,2$ (the order as in $\Omega$) person pick their pair of shoes, whereas $3,4$ doesn't pick their pair of shoes. There is simply $1$ way for person $1,2$ to pick their pair. So now, we're left with $10$ shoes (exactly $5$ pairs), and $2$ people to choose $2$ shoes each (in those $10$ shoes there are pairs of person $3$ and $4$). It will be easier, to count ways when at least one of them do select their pair. By Inclusion-Exlusion, from all possible ways ${10\choose 2}{8\choose 2}$, we need to substract $2 {8 \choose 2} - 1$ (where ${8 \choose 2}$ is the ways to choose any $2$ shoes out of $8$ shoes, when one person already selected his pair (we count it double, because it can be either person $1$ or $2$), and that $-1$ is due to possibility that both $1,2$ choose their pair.)

So that $|E| = {4 \choose 2}( {10 \choose 2}{8 \choose 2} - 2{8 \choose 2} + 1)$

And we have $\mathbb P(E) = \frac{{4 \choose 2}( {10 \choose 2}{8 \choose 2} - 2{8 \choose 2} + 1)}{{14 \choose 2}{12 \choose 2}{10 \choose 2}{8 \choose 2}}$