Poisson processes and ballistic missiles 
In response to an attack of 10 missiles, a response of 500 antiballistic missiles are launched. The missile targets of each antimissile are independent, and equiprobably picked. Each antimissile has a $.1$ chance to hit. What is the probability of all 10 missiles being hit?

The question then asks the student to use a Poisson process, although I'm willing to solve this using other methods. I thought I'd have
$$X=\sum X_i$$
with
$$X_i=1,0$$
if hit or not hit respectively. Then
$$P(X_i=1)=\sum_{k=1}^{500}\,_{500}\text{C}_kp^k(1-p)^{500-k}$$
and $p=1/100$.
I'm fairly certain my approach is wrong, and I do not know how to finish the problem.
 A: Exact Answer
For the exact answer, I would approach this using Inclusion-Exclusion.
Let $N_j$ be the event that incoming missile $j$ is missed.
The intersection of $k$ of the $N_j$ is $k$ particular missiles being missed. There is a $\frac{k}{10}$ chance that any will be targeted by a given anti-missile and a $\frac1{10}$ chance that that anti-missile will hit. Thus, there is a $\left(1-\frac{k}{10}\cdot\frac1{10}\right)$ chance that that anti-missile will miss, and a $\left(1-\frac{k}{100}\right)^{500}$ chance that all $500$ will miss. There are $\binom{10}{k}$ ways to choose the $k$ particular missiles Thus, the sum of the probabilities of $k$ missiles being missed by $500$ anti-missiles is
$$
\binom{10}{k}\left(1-\frac{k}{100}\right)^{500}
$$
Inclusion-Exclusion says that the probability of some missile not being hit is
$$
\sum_{k=1}^{10}(-1)^{k-1}\binom{10}{k}\left(1-\frac{k}{100}\right)^{500}=0.0638876
$$
Thus, the probability of all $10$ missiles being hit is $0.9361124$

Poisson Approximation
Since there are $500$ anti-missiles, each with a probability of $\frac1{100}$ of targeting a particular incoming missile, the expected number of hits on each missile is $5$. Thus, the Poisson approximation for getting no hits on a particular incoming missile is $e^{-5}$. Therefore, the probability of getting at least one hit on each particular missile is $\left(1-e^{-5}\right)$ and the probability of hitting all $10$ is
$$
\left(1-e^{-5}\right)^{10}=0.9346272
$$
This is pretty close to the exact answer.
A: This is problem 4.68 in "A first course in probability, 9th edition" by Ross. The back of the book claims that the answer is $\exp(-10e^{-5})  \approx 0.93484$. Let's reverse engineer that:
First, consider the 10 missiles and examine one of them, let's call it $M$. Let $X$ be the random variable that gives the number of anti-missiles that will hit $M$. Since each anti-missile has a $0.1 \times \frac{1}{10} = 0.01$ chance of hitting $M$, we see that $X$ is approximately Poisson with mean $500 \times 0.01 = 5$. Hence, the probability that no anti-missiles will hit $M$ is $e^{-5}$.
Next, let $Y$ be the random variable that gives the number of missiles that have no anti-missiles hitting them. This is approximately Poisson with mean $10 \times e^{-5}$. Hence the probability that all of the 10 missiles will have at least one anti-missile hitting it is equivalent to the the probability that none of the 10 missiles are unhit. This is just $\exp(-10e^{-5})$, QED.
