Suppose there were m married couples, but d of these 2m people have died. Regard the d deaths as striking the 2m people at random. Let X be the number of surviving couples.
Find:
a) E(X)
b) Var(X)
For part (a) I got E(X)=m((2m-d)/2m)^2 but im not sure how to calculate the variance.
The probability that a person survives is $1-\frac{d}{2m}$. Conditionally on the event that his person survives, the probability that a second person survives is $1-\frac{d}{2m-1}$.
Thus the probability that a given couple survives is $p=\left(1-\frac{d}{2m}\right)\cdot\left(1-\frac{d}{2m-1}\right)$ and the mean number of surviving couples is $mp=m\cdot\left(1-\frac{d}{2m}\right)\cdot\left(1-\frac{d}{2m-1}\right)$.
Let $U_c$ denote the indicator function of the event that couple $c$ survives. The reasoning above identifies $p=E[U_c]$ as $p=\left(1-\frac{d}{2m}\right)\cdot\left(1-\frac{d}{2m-1}\right)$ for every $c$ and uses the fact that $X=\sum\limits_{c=1}^mU_c$ to deduce that $E[X]=mp$. Let us now adapt this approach to the computation of the second moment of $X$.
For two different couples $b\ne c$, $U_bU_c$ is the indicator function of the event that both couples $c$ and $b$ survive. This has probability $q=\left(1-\frac{d}{2m}\right)\cdot\left(1-\frac{d}{2m-1}\right)\cdot\left(1-\frac{d}{2m-2}\right)\cdot\left(1-\frac{d}{2m-3}\right)$. Expanding the square $X^2$, one gets $m$ squares $U_c^2$, each such that $E[U_c^2]=E[U_c]=p$, and $m(m-1)$ products $U_cU_b$ with $b\ne c$, each such that $E[U_bU_c]=q$.
Hence $E[X^2]=mp+m(m-1)q$ and $\mathrm{var}(X)=E[X^2]-E[X]^2=mp+m(m-1)q-m^2p^2$.