# Numbers too large for my calculator, probability. Old calculator or should it be done another way?

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
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

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 $$\log x_k\approx (1000+\frac{1}{2})\log 1000-(k+\frac{1}{2}) \log k - (1000-k+\frac{1}{2})\log (1000-k)$$ $$-\frac{1}{2}\log 2\pi+k \log 0.2 + (1000-k) \log 0.8$$
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