Probability that a party with 20% support will get at most 14% of votes A certain political party has 20% support in a country with 10'000 citizens. What is the probability that at most 1'000 citizens will vote for that party assuming that 70% of the population is going to vote?
My solution: out of 2'000 people who support the party, pick $k$ randomly. Out of the rest pick $7000-k$. We are looking for the value of
$$\sum_{k=0}^{1000} \frac{{2000 \choose k} {8000 \choose 7000-k}}{10000 \choose 7000} = ?$$
According to Pari-GP, that's approximately $2.30366 \cdot 10^{-99}$. I have two questions:

*

*is my approach correct?

*how to approximate the probability without computer's aid? I was thinking about Lindenberg theorem but am not sure how to apply it.

 A: As Henry noted in a comment, two-thirds of the probability is in the contribution from $k=1000$. Note that as you decrease $k$ from this value, with each step you multiply the second factor roughly by $\frac{2000}{6000}=\frac13$, while the first factor stays roughly the same. Thus you get approximately a geometric series that sums to
$$
\frac{\binom{2000}{1000}\binom{8000}{6000}}{\binom{10000}{7000}}\cdot\frac1{1-\frac13}=\frac{\binom{2000}{1000}\binom{8000}{6000}}{\binom{10000}{7000}}\cdot\frac32\approx2.3065\cdot10^{-99}\;.
$$
If you regard computing the first factor with the binomial coefficients as “computer aid”, you can approximate them using Stirling’s approximation, which is applied to binomial coefficients in this Wikipedia section (which also provides a more specific approximation for the central binomial coefficient $\binom{2000}{1000}$).
A: Here is an alternate interpretation of the problem that leads to a different result, although still a tiny probability.
One interpretation of "20% support" is that the probability a randomly selected citizen will vote Yes is $0.2$.  Note that this is different than the assumption in the OP that exactly 2,000 of the 10,000 citizens will vote Yes if they vote at all.  With the alternate interpretation, the total number of Yes votes among the 7000 has a Binomial distribution with parameters $p=0.2$ and $n=7000$.  The Binomial can be approximated by a Normal distribution with mean $7000 \times 0.2 = 1400$ and variance $7000 \times 0.2 \times 0.8 = 1120$.  The resulting probability is too tiny to find in most tables, but using R, the probability that at most 1000 voters will votes Yes is then
$$pnorm(1000, \text{mean}=1400, \text{sd}=sqrt(1120)) = 3.16 \times 10^{-33}$$
Edit:
But see joriki's comment below--this is not a really good place to use the Normal approximation (although all methods agree that the probability is really tiny).
