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 $\colorbox[rgb]{1,1,1}{$.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:
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.
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 $\colorbox[rgb]{1,1,1}{$P$}\colorbox[rgb]{1,1,1}{$($}\colorbox[rgb]{1,1,1}{$A$}\colorbox[rgb]{1,1,1}{$B$}\colorbox[rgb]{1,1,1}{$)$}\colorbox[rgb]{1,1,1}{$=$}\colorbox[rgb]{1,1,1}{$P$}\colorbox[rgb]{1,1,1}{$($}\colorbox[rgb]{1,1,1}{$A$}\colorbox[rgb]{1,1,1}{$)$}\colorbox[rgb]{1,1,1}{$P$}\colorbox[rgb]{1,1,1}{$($}\colorbox[rgb]{1,1,1}{$B$}\colorbox[rgb]{1,1,1}{$)$}$ we just multiply these probabilities together.
What’s the probability that a missile doesn’t get through? It’s $\colorbox[rgb]{1,1,1}{$1$}\colorbox[rgb]{1,1,1}{$-$}\colorbox[rgb]{1,1,1}{$.05$}\colorbox[rgb]{1,1,1}{$=$}\colorbox[rgb]{1,1,1}{$.95$}$. Multiplying this number together 300 times gives ${\colorbox[rgb]{1,1,1}{$.95$}}^{\colorbox[rgb]{1,1,1}{$300$}}$.
So putting this together, the probability of at least one missile getting through is $\colorbox[rgb]{1,1,1}{$1$}\colorbox[rgb]{1,1,1}{$-$}{\colorbox[rgb]{1,1,1}{$.95$}}^{\colorbox[rgb]{1,1,1}{$300$}}\colorbox[rgb]{1,1,1}{$=$}\colorbox[rgb]{1,1,1}{$0.99999979246$}$. Notice this number is $$ but pretty darned close to it.
1. Missile Defense Version 2.0
What happens if you change .95 to .99 so it’s quite a bit more reliable. What’s the probability of New York being destroyed now? Answer: $\colorbox[rgb]{1,1,1}{$.951$}$
2. Dating 101
(a) Suppose when you throw a die the probability of getting any of the 6 possible outcomes is the same, as usual. Now suppose you throw this die 10 times. What’s the probability that you’ll never get a 1?
Now let’s try to apply this to a real life situation. Bob wants to go out on a date. His roommate Jim, says that about one in every 5 times he asks someone out, they agree. So Bob decides to ask Mary out. He figures he’ll ask her out 25 times. Well what’s the probability of him getting at least one date with her? Using the same logic as before, you see the probability is $\colorbox[rgb]{1,1,1}{$1$}\colorbox[rgb]{1,1,1}{$-$}{\colorbox[rgb]{1,1,1}{$.8$}}^{\colorbox[rgb]{1,1,1}{$25$}}\colorbox[rgb]{1,1,1}{$=$}\colorbox[rgb]{1,1,1}{$.996$}$ which is pretty good odds.
As a bit of common sense will tell you, this tactic won’t work. If Mary says no to Bob, the repetition of the same offer in rapid succession 25 times will only irritate her, reinforcing in her mind the opinion that a date with Bob would be rather less pleasant than a tooth extraction. If Bob wants to go out on dates, he should try changing majors and become a lawyer.
Bob is not only socially challenged, but he should have also realized that applying a formula is only expected to work if the conditions on its derivation are satisfied.
(b) What are the conditions in the derivation that weren’t satisfied?