Expected number of remaining pairs of animals Suppose that Noah started with $n$ pairs of animals on the ark and that $m$ animals died. If the $m$ animals were chosen randomly, what is the expected number of complete pairs left?
My attempt: Total animals: $2n$
Probability of dying: $\frac{m}{2n}$
The probability that a pair $X_i$ is left intact is $$\left(1-\frac{m}{2n}\right)^2,$$ so the expected number of remaining pairs is $$n\left(1-\frac{m}{2n}\right)^2$$
Am I correct?
 A: The probability for both giraffes to survive is the probability that the first death is not a giraffe, multiplied by the probability that the second death is not a giraffe, and so on until the probability that the $m$-th death is not a giraffe:
$$P(\hbox{lucky giraffes})=\frac{2n-2}{2n}.\frac{2n-3}{2n-1}\cdots\frac{2n-m-1}{2n-m+1}=\frac{(2n-m)(2n-m-1)}{2n(2n-1)}$$
This is, then, the expected number of giraffe couples to survive. By symmetry it is also the expected number of platypus couples to survive, and so on. The expected total number of couples to survive is then $n$ times that number.
Disclosure: I borrowed heavily from this question although I do not understand the complete answer there!
A: For a given pair, there are $\binom{2n-2}{m}$ arrangements of $m$ deaths that leave that pair alone. There are $\binom{2n}{m}$ arrangements of $m$ deaths. Thus, the probability that a given pair survives is
$$
\frac{\binom{2n-2}{m}}{\binom{2n}{m}}
$$
Thus, by linearity of expectation, the expected number of surviving pairs is
$$
n\frac{\binom{2n-2}{m}}{\binom{2n}{m}}
$$
A: Your answer for the probability that a pair $X_{i}$ is left intact requires a small correction since the probabilities are not independent.
The probability that the first animal in the pair is left intact is $\left(1-\frac{m}{2n}\right)$.
Now, the probability that the second animal in the pair is $\left(1-\frac{m}{2n-1}\right)$ since we know now that the first one was already intact.
Now, Multiplying these gives the intersection, i.e. $$Pr[X_{i}\text{ is left intact}]=\left(\frac{(2n-m)(2n-1-m)}{(2n-1)(2n)}\right)$$
Which is the same as in the combinatorial argument.
Now the expectation is n times the probability, as explained above.
