Probability question...

Question

In a survey, N married couples are observed (i.e. N men and N women). A few decades later, in a follow-up survey, it was discovered that k individuals, among those who were surveyed, have died. Assuming that the deaths were random, the expected number of couple who survive is

Answer 1

The answer is $\newcommand{\chh}[2]{\left(\begin{array}{c}#1\\#2\end{array}\right)}N\frac{\chh{2N-2}{k}}{\chh{2N}{k}}=\frac{(2N-k)(2N-k-1)}{2(2N-1)}$.

Proof)

Let $X_i$ for $i=1,\ldots,N$ denote whether or not $i$th couple has survived. That is

$$ X_i = \left\{\begin{array}{ll} 1 & \mbox{if $i$th couple has survived} \\ 0 & \mbox{otherwise} \end{array}\right. $$

Then the expected number of couple who survive is $$ \mathbf{E} \left(\sum_{i=1}^N X_i \right) = \sum_{i=1}^N \mathbf{E} (X_i). $$

This is true whether or not $X_i$ are independent since the expected value is a linear operator.

Now $\mathbf{E} X_i$ stands for the probability that $i$th couple has survived and $$ \frac{\chh{2N-2}{k}}{\chh{2N}{k}} $$ since the total number of ways that one can choose $k$ people out of $2N$ people is $\chh{2N}{k}$ and the total number of ways that one can choose $k$ people out of $2N-2$ people, i.e., everyone but this couple, is $\chh{2N-2}{k}$.

Note that $\chh{n}{k}$ refers to the number of ways of choosing $k$ items out of $n$ items, i.e., $$ \chh{n}{k} = \frac{n!}{k!(n-k)!}. $$

Therefore, the expected number of couple who survive is $$ \mathbf{E} \left(\sum_{i=1}^N X_i \right) = \sum_{i=1}^N \mathbf{E} (X_i) = N \mathbf{E} X_1 = N\frac{\chh{2N-2}{k}}{\chh{2N}{k}} =\frac{(2N-k)(2N-k-1)}{2(2N-1)}. $$

Answer 2

The probability of a specific woman surving is $1$ minus the probability of her dying and the probability of her dying is $\frac k{2N}$. So the probability of a her living is $1 -\frac k{2N}$.

So given a married couple the probability of the wife surviving is $(1-\frac k{2N})$. Given that of the remaining $2N-1$ people $k$ have died the conditional probability of her husband surviving given that the wife did, is $1-\frac k{2N-1}$ and the probability of a couple surving is $(1-\frac k{2N})(1-\frac k{2N-1})$.

And so of the $N$ married couples you will expect $N(1-\frac k{2N})(1-\frac k{2N-1})$ to still be intact.

The expression $N(1-\frac k{2N})(1-\frac k{2N-1})$ can be rewritten as $N(\frac {2N -k}{2N})(\frac {2N-k-1}{2N-1}) = \frac {(2N-k)(2N-k-1)}{2(2N-1)}$ but I don't know if that makes things better.

Credit to saulspatz for noticing two mistakes I made in my previous answer.