Stuck on a probability question about pairs of socks

combinatoricsprobability

The question is as follows:

You have 10 pairs of socks (i.e., 20 socks in total) with each pair in a different color. You put all the socks into the washing machine but it “eats” four of the 20 socks at random.

What is the expected number of complete pairs left in the washing machine?

I approached this question like this:

There are 3 possible outcomes:

  1. 8 pairs of socks survive
  2. 7 pairs of socks survive
  3. 6 pairs of socks survive

I then tried to find the probability of each of these events occurring.

The ways that 8 pairs of socks survive is:

  1. Eating 2 different colour socks then eating their pairs

  2. Eating first sock, then eating it's pair, Eating the second sock, then eating its pair

P(8 pairs) = $1*\frac{18}{19}*\frac{2}{18}*\frac{1}{17}\\+1*\frac{1}{19}*1*\frac{1}{17}$

The ways that 7 pairs of socks survive is:

  1. Eating three socks of different colours, then eating any one of their pairs
  2. Eating two socks of different colours, then eating any one of their pairs, then drawing another sock of a different colour
  3. Eating any sock, eating another sock of the same colour, then eating two other socks of different colours

P(7 socks)=$1*\frac{18}{19}*\frac{16}{18}*\frac{3}{17}\\+1*\frac{18}{19}*\frac{2}{18}*\frac{16}{17}\\+1*\frac{1}{19}*1*\frac{16}{17}$

The ways that 6 pairs of socks survive is:

  1. Eating 4 socks of different colours

P(6 pairs)=$1*\frac{18}{19}*\frac{16}{18}*\frac{14}{17}$

I am confident that my solution is incorrect.

First of all, the probabilities sum to $\frac{2906}{2907}$ when it should sum to 1.

Secondly, I simulated the process in the question on my computer and the probabilities that I obtained were:

P(8 pairs)=$0.0269986$

P(7 pairs)=$0.4319539$

P(6 pairs)=$0.5410475$

Compared to my answers which were:

P(8 pairs)=$0.0093$

P(7 pairs)=$0.2972$

P(6 pairs)=$0.6935$

Would there be another simpler way of solving the question?

Best Answer

There are $\binom{20}{16}$ ways to select the $16$ remaining socks, all of which we assume are equally likely. Let $$X_i = \begin{cases} 1 \qquad \text{if pair i survives intact} \\ 0 \qquad \text{otherwise} \end{cases}$$ for $1 \le i \le 10$. So $$P(X_i = 1) = \frac{\binom{18}{14}}{\binom{20}{16}} =\frac{3060}{4845}$$ By linearity of expectation, the expected value of the number of surviving pairs is $$E \left( \sum_{i=1}^{10} X_i \right) = \sum_{i=1}^{10} E(X_i) = 10 \times \frac{3060}{4845} = \boxed{6.31579}$$

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