1.3.6 Return to New York

Now let’s go back to the star wars missile defense example we talked about before in section 1.3 and see if we can answer the question.

We’re assuming you’ve got 300 missiles. The probability of any one getting through is $.05$ and we’re assuming independence of the different missiles. That is, the outcome of one getting through doesn’t influence any of the others. So this is how you can solve this problem:

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  1. What’s the probability that at least one gets through?

  Well the are a lot of possibilities. The first and third one could get through. Or the 3rd, 5th, and 101st could get through. This is a lot harder to enumerate than the dice example, but we don’t have to. We just notice that at least one getting through has all possibilities, except one: none getting through. Therefore the probability the at least one gets through is 1 minus the probability that none get through.

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  2. What’s the probability that none get through?

  That’s a lot easier to calculate because that can only happen one way. All the missiles have to miss. So the probability that none get through is the probability (the first one doesn’t make it) and (the second one doesn’t) and (the 3rd one doesn’t) … and (the last one doesn’t). But these are all independent. So since we know in this case $P(AB)=P(A)P(B)$ we just multiply these probabilities together.

  What’s the probability that a missile doesn’t get through? It’s $1-.05=.95$. Multiplying this number together 300 times gives ${.95}^{300}$.

So putting this together, the probability of at least one missile getting through is $1-{.95}^{300}=0.99999979246$. Notice this number is but pretty darned close to it.