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It's basic binomial probability, the problem is that the numbers are to large for my calculator. Is this the point of the task, to force me to use another method? It doesn't seem lightly, because n over r is a must anyways. An example would be 1000 over 180.

Note that on the exam it self I am allowed to use my laptop (mac), but not the net. So if you can think of any programs that can deal with this size of numbers I'd appreciate it.

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You should explain more about what the task is. How are we supposed to figure out what the goal of the assignment was without knowing what was assigned? –  Zev Chonoles Jul 22 '11 at 7:33
    
The problem applies to a lot of different tasks where the numbers are to big, so it really is in general. –  Algific Jul 22 '11 at 7:36
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Is the problem you are solving related to the binomial distribution? If so, there is the normal approximation(en.wikipedia.org/wiki/…) and the Poisson approximation (en.wikipedia.org/wiki/…). I am suggesting this in a comment because it doesn't directly answer your question as stated. –  Henry B. Jul 22 '11 at 7:46

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You can use Stirling's approximation to get the logs of the factorials. So if $$x_k={1000\choose k}0.2^k0.8^{1000-k}=\frac{1000!}{k!(1000-k)!}0.2^k0.8^{1000-k}$$ you have $$\log x_k\approx (1000+\frac{1}{2})\log 1000-(k+\frac{1}{2}) \log k - (1000-k+\frac{1}{2})\log (1000-k)$$ $$ -\frac{1}{2}\log 2\pi+k \log 0.2 + (1000-k) \log 0.8$$

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Since I'm allowed to use my computer the bcd function on my casio can be done in excel with the following line: =BINOMDIST(180,1000,0.2,TRUE). The true part is because we want the probability for at most 180 people/success.

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Well, problems like this are the reason, why the books and courses teach about the normal approximation! Frankly, it is hard to believe that the teacher would want you to compute all 181 terms by frantically punching a calculator. I suggest that for preparatory problems you: 1) try more realistic numbers, 2) learn about the normal approximation. Presumably you can use a table or something to produce the cdf of a normally distributed random variable? –  Jyrki Lahtonen Jul 22 '11 at 10:37
    
But you probably can (in a way) do it with your calculator by starting with a term $x_k={1000\choose k}0.2^k0.8^{1000-k}$ that your calculator can do accurately (relatively small $k$). Then you can use the formula $x_{k+1}=x_k*(1000-k)*0.2/(0.8*(k+1))$. And then in the end sum $x_0+x_1+\cdots+x_{180}$. But I really don't think that you want to do that, and numerical inaccuracies are bound to make this doubtful anyway. –  Jyrki Lahtonen Jul 22 '11 at 10:51
    
I got in contact with the author and we are allowed to use laptops on the exam. Problem solved. –  Algific Sep 10 '11 at 18:54

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