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.

• I have seen this question before on this site. I don't have the link at the moment. – Chinny84 Mar 11 '15 at 23:49

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.

• These answers makes sense to me. I appreciate the thoroughness. Thanks! – ok_ Mar 12 '15 at 2:01
• @robjohn Could one also say that the probability a particular missile is hit is $(1-(1-0.01)^{500})^{10}=0.9362042...$? Or is this wrong? It seems to be really close to the answer, is this just a fluke? – Guacho Perez Dec 8 '17 at 2:55