Rrhain writes:
quote:
Here's the general solution for n, just do the math to find the answers for 10, 20 and ∞ darts:
Pn = 1 - (1 - 1/n)n
You need to show your work.
This particular part is too elemenary for that, I just wrote it down off the top of my head. I could break it down for you, but why don't we let someone who's figuring it out for the first time do that? This isn't the interesting part of the problem anyway.
Yes, that is the correct formula, but you neglected to say why.
The original hint you provided explains why.
quote:
(1 - (1 - 1/n)n) = .01
That's the answer, but I don't know how to solve the equation.
Incorrect. That is not the answer. The term for
∞ does not go to 0.
Oh, you're right! Isn't that interesting. I still don't know how to solve the equation. For example, I don't know how to find log(1-1/n). Just by inspection it looks like n is less than 20, though.
You originally introduced the problem in a reply to Daddy's
Message 38 where he was talking about zero chances in a kajilion. Can you tie your probability problem back into the misconception Daddy was experiencing? I think that would be pretty helpful.
--Percy
PS - Your problem showed up a bug in MathCad, which believes the expression for P
∞ goes to 0.